Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices Miscellaneous Solutions

(I) Choose The Correct Alternative.

Question 1. If AX = B, where \( A = \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \) then X =
(a) \( \begin{bmatrix} 3/5 \\ 3/7 \end{bmatrix} \)
(b) \( \begin{bmatrix} 7/3 \\ 5/3 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 \\ 2 \end{bmatrix} \)
Answer: (c) \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)
In simple words: To find X, we need to solve the matrix equation AX=B by first finding the inverse of A, then multiplying A-1 by B. For this specific problem, \( A^{-1} = \frac{1}{(-1)(-1) - (2)(2)} \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} = \frac{1}{1-4} \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} = \begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix} \). Then \( X = A^{-1}B = \begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} (1/3)(3) + (2/3)(1) \\ (2/3)(3) + (1/3)(1) \end{bmatrix} = \begin{bmatrix} 1 + 2/3 \\ 2 + 1/3 \end{bmatrix} = \begin{bmatrix} 5/3 \\ 7/3 \end{bmatrix} \). Oh, wait, the provided answer is \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \). Let's recheck the calculation for A inverse and X: For \( A = \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix} \), \( det(A) = (-1)(-1) - (2)(2) = 1 - 4 = -3 \). \( adj(A) = \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} \). \( A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} = \begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix} \). Then \( X = A^{-1} B = \begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} (1/3)(3) + (2/3)(1) \\ (2/3)(3) + (1/3)(1) \end{bmatrix} = \begin{bmatrix} 1 + 2/3 \\ 2 + 1/3 \end{bmatrix} = \begin{bmatrix} 5/3 \\ 7/3 \end{bmatrix} \). The options provided do not match my calculated X. However, the selected answer (c) is \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \). Let's check if \( A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = B \). \( \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} (-1)(1) + (2)(1) \\ (2)(1) + (-1)(1) \end{bmatrix} = \begin{bmatrix} -1 + 2 \\ 2 - 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \). This result is \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \), but B is \( \begin{bmatrix} 3 \\ 1 \end{bmatrix} \). So, option (c) is incorrect based on the provided A and B. There seems to be an error in the question or the given answer. Assuming the provided answer (c) is intended to be correct, there might be a typo in the original question's matrix A or B. However, following the instruction to output verbatim, I will provide the answer as given. The matrix X is found by computing the inverse of matrix A and then multiplying it by matrix B. The key is to correctly perform matrix inversion and multiplication.

🎯 Exam Tip: For matrix equations like AX=B, remember that \( X = A^{-1}B \). This requires calculating the determinant and adjoint of A to find its inverse, followed by matrix multiplication. Be precise with calculations as small errors can lead to incorrect options.

Question 2. The matrix \( \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} \) is __________
(a) identity matrix
(b) scalar matrix
(c) null matrix
(d) diagonal matrix
Answer: (b) scalar matrix
In simple words: A scalar matrix is a diagonal matrix where all the diagonal elements are equal to each other. Since all diagonal elements are 8 and off-diagonal elements are 0, it's a scalar matrix.

🎯 Exam Tip: Understand the definitions of different types of matrices (identity, scalar, null, diagonal) to quickly identify them. A scalar matrix is a special type of diagonal matrix where all diagonal entries are identical non-zero values.

Question 3. The matrix \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \) is __________
(a) identity matrix
(b) diagonal matrix
(c) scalar matix
(d) null matrix
Answer: (d) null matrix
In simple words: A null matrix (or zero matrix) is a matrix where all its elements are zero. This matrix clearly has all elements as zero.

🎯 Exam Tip: The null matrix is a fundamental matrix where every element is zero, serving as the additive identity in matrix addition. Confidently distinguishing it from other matrix types is crucial.

Question 4. If \( A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix} \), then \( |\text{adj} A| = \) __________
(a) \( a^{12} \)
(b) \( a^9 \)
(c) \( a^6 \)
(d) \( a^{-3} \)
Answer: (c) \( a^6 \)
Hint: \[ \text{adj} A = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 \end{bmatrix} \] \( \therefore |\text{adj} A| = a^2(a^4 - 0) = a^6 \)
In simple words: For a scalar matrix A of order n, \( A = kI \), where I is the identity matrix. Its determinant is \( |A| = k^n \). The formula for the determinant of the adjoint of a matrix is \( |\text{adj} A| = |A|^{n-1} \). In this case, A is a 3x3 scalar matrix with \( k=a \), so \( |A| = a^3 \). Therefore, \( |\text{adj} A| = (a^3)^{3-1} = (a^3)^2 = a^6 \).

🎯 Exam Tip: Remember the property that for an n x n matrix A, \( |\text{adj} A| = |A|^{n-1} \). For a diagonal matrix, the determinant is the product of its diagonal elements. For a scalar matrix \( A = kI \), \( |A| = k^n \).

Question 5. Adjoint of \( \begin{bmatrix} 2 & -3 \\ 4 & -6 \end{bmatrix} \) is __________
(a) \( \begin{bmatrix} -6 & 3 \\ -4 & 2 \end{bmatrix} \)
(b) \( \begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix} \)
(c) \( \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \)
(d) \( \begin{bmatrix} 4 & 2 \\ -6 & -3 \end{bmatrix} \)
Answer: (a) \( \begin{bmatrix} -6 & 3 \\ -4 & 2 \end{bmatrix} \)
In simple words: For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), its adjoint is found by swapping the diagonal elements (a and d) and changing the signs of the off-diagonal elements (b and c). So for \( \begin{bmatrix} 2 & -3 \\ 4 & -6 \end{bmatrix} \), swap 2 and -6, change signs of -3 and 4, which gives \( \begin{bmatrix} -6 & -(-3) \\ -(4) & 2 \end{bmatrix} = \begin{bmatrix} -6 & 3 \\ -4 & 2 \end{bmatrix} \).

🎯 Exam Tip: The shortcut for finding the adjoint of a 2x2 matrix is crucial for speed and accuracy. Swap the main diagonal elements and negate the off-diagonal elements. This is a common exam calculation.

Question 6. If A = diag. \( [d_1, d_2, d_3, ..., d_n] \), where \( d_i \neq 0 \), for i = 1, 2, 3, ......., n, then \( A^{-1} = \) __________
(a) diag \( [1/d_1, 1/d_2, 1/d_3, ..., 1/d_n] \)
(b) D
(c) I
(d) O
Answer: (a) diag \( [1/d_1, 1/d_2, 1/d_3, ..., 1/d_n] \)
In simple words: For a diagonal matrix, its inverse is also a diagonal matrix where each diagonal element is the reciprocal of the corresponding diagonal element in the original matrix. This property only holds if none of the diagonal elements are zero.

🎯 Exam Tip: The inverse of a diagonal matrix is straightforward: simply take the reciprocal of each diagonal element. This property greatly simplifies inverse calculations for such matrices, provided all diagonal elements are non-zero.

Question 7. If \( A^2 + mA + nI = O \) and \( n \neq 0, |A| \neq 0 \), then \( A^{-1} = \) __________
(a) \( -\frac{1}{m}(A + nI) \)
(b) \( -\frac{1}{n}(A + mI) \)
(c) \( -\frac{1}{m}(I + mA) \)
(d) \( (A + mnI) \)
Answer: (b) \( -\frac{1}{n}(A + mI) \)
Hint: \( A^2 + mA + nI = O \)
\( \therefore (A^2 + mA + nI) . A^{-1} = O . A^{-1} \)
\( \therefore A(AA^{-1}) + m(AA^{-1}) + nIA^{-1} = O \)
\( \therefore AI + mI + nA^{-1} = O \)
\( \therefore nA^{-1} = -A - mI \)
\( \therefore A^{-1} = -\frac{1}{n}(A + mI) \)
In simple words: To find the inverse of A from the given equation, multiply the entire equation by \( A^{-1} \) from the right. Remember that \( A A^{-1} = I \) and \( I A^{-1} = A^{-1} \). Then, rearrange the terms to isolate \( A^{-1} \).

🎯 Exam Tip: When deriving matrix inverses from characteristic equations, multiply the entire equation by \( A^{-1} \). This technique is fundamental for solving matrix algebra problems involving polynomial expressions of matrices.

Question 8. If a 3 x 3 matrix B has its inverse equal to B, then \( B^2 = \) __________
(a) \( \begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \)
(b) \( \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Answer: (d) \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Hint: \( B^{-1} = B \)
\( \therefore B^2 = B . B^{-1} = I \)
In simple words: If a matrix B is its own inverse, it means that when you multiply B by itself, the result is the identity matrix (I). This is a property of involutory matrices.

🎯 Exam Tip: The definition of an inverse matrix is \( B B^{-1} = I \). If \( B^{-1} = B \), then \( B^2 = I \), where I is the identity matrix of the same order as B. Recognize the identity matrix, which has 1s on the main diagonal and 0s elsewhere.

Question 9. If \( A = \begin{bmatrix} \alpha & 4 \\ 4 & \alpha \end{bmatrix} \) and \( |A^3| = 729 \) then \( \alpha = \) __________
(a) \( \pm 3 \)
(b) \( \pm 4 \)
(c) \( \pm 5 \)
(d) \( \pm 6 \)
Answer: (c) \( \pm 5 \)
Hint: \( |A| = \begin{vmatrix} \alpha & 4 \\ 4 & \alpha \end{vmatrix} = \alpha^2 - 16 \)
\( \therefore |A^3| = |A|^3 = (\alpha^2 - 16)^3 = 729 \)
\( \therefore \alpha^2 - 16 = 9 \)
\( \therefore \alpha^2 = 25 \)
\( \therefore \alpha = \pm 5 \)
In simple words: The determinant of a matrix raised to a power is equal to the determinant of the matrix raised to that power, i.e., \( |A^n| = |A|^n \). First, calculate the determinant of A, then set \( |A|^3 \) equal to 729 and solve for \( \alpha \).

🎯 Exam Tip: Remember the property \( |A^n| = |A|^n \). This simplifies calculations immensely. Also, recall how to find the determinant of a 2x2 matrix and solve basic algebraic equations for unknown variables.

Question 10. If A and B square matrices of order n x n such that \( A^2 - B^2 = (A - B)(A + B) \), then which of the following will be always true?
(a) AB = BA
(b) either A or B is a zero matrix
(c) either of A and B is an identity matrix
(d) A = B
Answer: (a) AB = BA
Hint: \( A^2 - B^2 = (A - B)(A + B) \)
\( \therefore A^2 - B^2 = A^2 + AB - BA - B^2 \)
\( \therefore O = AB - BA \)
\( \therefore AB = BA \)
In simple words: For the algebraic identity \( (A-B)(A+B) = A^2 - B^2 \) to hold true for matrices, it is necessary that the matrices A and B commute, meaning their multiplication order does not affect the result (AB = BA). If they don't commute, \( AB - BA \neq O \).

🎯 Exam Tip: Unlike scalar algebra, matrix multiplication is generally not commutative (\( AB \neq BA \)). The identity \( (A-B)(A+B) = A^2 - B^2 \) holds true if and only if matrices A and B commute (i.e., AB = BA). This is a crucial distinction in matrix algebra.

Question 11. If \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \) then \( A^{-1} = \) __________
(a) \( \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)
(b) \( \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)
(c) \( \begin{bmatrix} 3 & 5 \\ -1 & 2 \end{bmatrix} \)
(d) \( \begin{bmatrix} 3 & -5 \\ 1 & -2 \end{bmatrix} \)
Answer: (b) \( \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)
In simple words: To find the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), use the formula \( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \). For matrix A, \( det(A) = (2)(3) - (5)(1) = 6 - 5 = 1 \). Then, \( A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \).

🎯 Exam Tip: Memorize the formula for the inverse of a 2x2 matrix. It is a frequent calculation. Ensure correct calculation of the determinant and sign changes for the off-diagonal elements. The determinant must be non-zero for the inverse to exist.

Question 12. If A is a 2 x 2 matrix such that \( A(\text{adj } A) = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \), then \( |A| = \) __________
(a) 0
(b) 5
(c) 10
(d) 25
Answer: (b) 5
Hint: \( A(\text{adj } A) = |A|.I \)
In simple words: The product of a matrix A and its adjoint (adj A) is always equal to the determinant of A multiplied by the identity matrix (I). Here, \( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \) can be written as \( 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), which means it's \( 5I \). Comparing this to \( |A|I \), we find that \( |A| = 5 \).

🎯 Exam Tip: A key property in matrix algebra is \( A(\text{adj } A) = (\text{adj } A)A = |A|I \). This identity allows for direct determination of the determinant \( |A| \) when the product \( A(\text{adj } A) \) is given. Recognize scalar multiples of the identity matrix.

Question 13. If A is a non-singular matrix, then \( \text{det}(A^{-1}) = \) __________
(a) 1
(b) 0
(c) det(A)
(d) 1/det(A)
Answer: (d) 1/det(A)
Hint: \( AA^{-1} = I \)
\( \therefore |A|.|A^{-1}| = |I| \)
\( \therefore |A^{-1}| = \frac{1}{|A|} \)
In simple words: The determinant of the inverse of a matrix is equal to the reciprocal of the determinant of the original matrix. This is because the product of a matrix and its inverse is the identity matrix, and the determinant of the identity matrix is 1.

🎯 Exam Tip: Understand the property \( |AB| = |A||B| \) and that \( |I| = 1 \). Applying these to \( AA^{-1} = I \) immediately yields \( |A^{-1}| = \frac{1}{|A|} \). This is fundamental for working with determinants and inverses.

Question 14. If \( A = \begin{bmatrix} 1 & 2 \\ -3 & 1 \end{bmatrix}, B = \begin{bmatrix} -1 & 0 \\ 1 & 5 \end{bmatrix} \) then AB = __________
(a) \( \begin{bmatrix} 1 & -10 \\ 1 & 20 \end{bmatrix} \)
(b) \( \begin{bmatrix} 1 & 10 \\ -1 & 20 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & 10 \\ 1 & 20 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 10 \\ -1 & -20 \end{bmatrix} \)
Answer: (c) \( \begin{bmatrix} 1 & 10 \\ 1 & 20 \end{bmatrix} \)
In simple words: To find the product AB, perform row-by-column multiplication. Multiply the rows of A by the columns of B. \( AB = \begin{bmatrix} (1)(-1) + (2)(1) & (1)(0) + (2)(5) \\ (-3)(-1) + (1)(1) & (-3)(0) + (1)(5) \end{bmatrix} \) \( = \begin{bmatrix} -1 + 2 & 0 + 10 \\ 3 + 1 & 0 + 5 \end{bmatrix} = \begin{bmatrix} 1 & 10 \\ 4 & 5 \end{bmatrix} \). Again, there seems to be a discrepancy between my calculation and the provided answer. The question says (c) \( \begin{bmatrix} 1 & 10 \\ 1 & 20 \end{bmatrix} \). If we try to work backwards, for the second row, \( (-3)(-1)+(1)(1) = 3+1 = 4 \) (not 1) and \( (-3)(0)+(1)(5) = 5 \) (not 20). There appears to be an error in the provided options or the selected answer. Based on standard matrix multiplication, the result is \( \begin{bmatrix} 1 & 10 \\ 4 & 5 \end{bmatrix} \). However, I will output the answer as given, i.e., option (c).

🎯 Exam Tip: Matrix multiplication requires careful calculation of each element by summing the products of corresponding elements from a row of the first matrix and a column of the second matrix. Pay close attention to signs and positions to avoid errors.

Question 15. If x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6, then (y, z) = __________
(a) (-1, 0)
(b) (1, 0)
(c) (1, -1)
(d) (-1, 1)
Answer: (b) (1, 0)
In simple words: This is a system of linear equations. You can solve it using methods like substitution, elimination, or matrix methods (inverse matrix method or Cramer's rule). Let's use elimination: 1. \( x + y + z = 3 \) 2. \( x + 2y + 3z = 4 \) 3. \( x + 4y + 9z = 6 \) Subtract (1) from (2): \( (x + 2y + 3z) - (x + y + z) = 4 - 3 \) \( y + 2z = 1 \) (Equation A) Subtract (1) from (3): \( (x + 4y + 9z) - (x + y + z) = 6 - 3 \) \( 3y + 8z = 3 \) (Equation B) Now we have a system with two variables: A. \( y + 2z = 1 \implies y = 1 - 2z \) B. \( 3y + 8z = 3 \) Substitute A into B: \( 3(1 - 2z) + 8z = 3 \) \( 3 - 6z + 8z = 3 \) \( 3 + 2z = 3 \) \( 2z = 0 \) \( z = 0 \) Substitute z = 0 back into \( y = 1 - 2z \): \( y = 1 - 2(0) \) \( y = 1 \) So, \( (y, z) = (1, 0) \). (We could also find x: \( x + 1 + 0 = 3 \implies x = 2 \).) The solution matches option (b).

🎯 Exam Tip: Systems of linear equations can be efficiently solved using elimination. Reduce the system to fewer variables step-by-step. Double-check your calculations, especially when substituting values back into equations.

(II) Fill In The Blanks:

Question 1. \( A = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \) is __________ matrix.
Answer: column
In simple words: A column matrix (or column vector) is a matrix that has only one column, regardless of the number of rows. This matrix has two rows and one column.

🎯 Exam Tip: Identify matrix types based on their dimensions. A matrix with only one column is a column matrix, and one with only one row is a row matrix.

Question 2. Order of matrix \( \begin{bmatrix} 2 & 1 & 1 \\ 5 & 1 & 8 \end{bmatrix} \) is __________
Answer: 2 x 3
In simple words: The order of a matrix is defined by its number of rows by its number of columns (rows x columns). This matrix has 2 rows and 3 columns.

🎯 Exam Tip: Always state the order of a matrix as "rows x columns". This convention is crucial for matrix operations like multiplication and understanding matrix dimensions.

Question 3. If \( A = \begin{bmatrix} 4 & x \\ 6 & 3 \end{bmatrix} \) is a singular matrix, then x is __________
Answer: 2
In simple words: A singular matrix is a square matrix whose determinant is zero. For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the determinant is \( ad-bc \). Set the determinant of A to zero and solve for x. \( (4)(3) - (x)(6) = 0 \) \( 12 - 6x = 0 \) \( 6x = 12 \) \( x = 2 \)

🎯 Exam Tip: The condition for a matrix to be singular is that its determinant must be zero. This is a fundamental concept for understanding matrix invertibility and solving related problems.

Question 4. Matrix \( B = \begin{bmatrix} 0 & 3 & 1 \\ -3 & 0 & -4 \\ p & 4 & 0 \end{bmatrix} \) is a skew-symmetric, then value of p is __________
Answer: -1
In simple words: A matrix B is skew-symmetric if its transpose is equal to the negative of the matrix, i.e., \( B^T = -B \). This implies that the elements satisfy \( b_{ij} = -b_{ji} \) for all i and j, and \( b_{ii} = 0 \) (diagonal elements are zero). From \( B^T = -B \), comparing elements: \( b_{31} = p \), and \( b_{13} = 1 \). So, \( p = -b_{13} = -1 \). Also check: \( b_{21} = -3 \), \( b_{12} = 3 \). \( -3 = -3 \). \( b_{32} = 4 \), \( b_{23} = -4 \). \( 4 = -(-4) \). This confirms p = -1.

🎯 Exam Tip: For a skew-symmetric matrix, remember two key properties: all diagonal elements are zero, and \( a_{ij} = -a_{ji} \) for \( i \neq j \). Use these properties to find unknown elements quickly.

Question 5. If A = \( [a_{ij}]_{2x3} \) and B = \( [b_{ij}]_{m \times 1} \), and AB is defined, then m = __________
Answer: 3
In simple words: For matrix multiplication AB to be defined, the number of columns in the first matrix (A) must be equal to the number of rows in the second matrix (B). A has order 2x3, so it has 3 columns. B has order mx1, so it has m rows. For AB to be defined, \( 3 = m \).

🎯 Exam Tip: The compatibility condition for matrix multiplication AB is that the number of columns in A must equal the number of rows in B. If A is \( p \times q \) and B is \( q \times r \), then AB is \( p \times r \).

Question 6. If \( A = \begin{bmatrix} 3 & -5 \\ 2 & 5 \end{bmatrix} \), then cofactor of \( a_{12} \) is __________
Answer: -2
In simple words: The cofactor of an element \( a_{ij} \) is given by \( C_{ij} = (-1)^{i+j} M_{ij} \), where \( M_{ij} \) is the minor of \( a_{ij} \). The element \( a_{12} \) is -5. Its minor \( M_{12} \) is the determinant of the matrix formed by removing the 1st row and 2nd column, which is just 2. So, \( C_{12} = (-1)^{1+2} M_{12} = (-1)^3 (2) = -1 \times 2 = -2 \).

🎯 Exam Tip: Remember the formula for cofactors: \( C_{ij} = (-1)^{i+j}M_{ij} \). Carefully determine the minor \( M_{ij} \) by deleting the corresponding row and column, and pay attention to the sign determined by \( (-1)^{i+j} \).

Question 7. If A = \( [a_{ij}]_{m \times m} \) is non-singular matrix, then \( A^{-1} = \frac{1}{\text{____}} \text{adj (A)} \).
Answer: \( |A| \)
In simple words: This is the standard formula for finding the inverse of a matrix using its adjoint. For a non-singular matrix A (where \( |A| \neq 0 \)), its inverse is obtained by dividing the adjoint of A by the determinant of A.

🎯 Exam Tip: The formula \( A^{-1} = \frac{1}{|A|} \text{adj (A)} \) is fundamental for calculating matrix inverses. It emphasizes that a matrix is invertible if and only if its determinant is non-zero (i.e., it is non-singular).

Question 8. \( (A^T)^T = \) __________
Answer: A
In simple words: The transpose of the transpose of a matrix A is the original matrix A itself. Transposing a matrix swaps its rows and columns; doing it twice brings it back to the original arrangement.

🎯 Exam Tip: Understand basic matrix properties like \( (A^T)^T = A \). These properties are often used in proofs and simplifying matrix expressions.

Question 9. If \( A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \) and \( A^{-1} = \begin{bmatrix} 1 & -1 \\ x & 2 \end{bmatrix} \), then x = __________
Answer: -1
In simple words: For a matrix and its inverse, their product is the identity matrix, i.e., \( AA^{-1} = I \). Let's calculate \( A^{-1} \) for A: \( |A| = (2)(1) - (1)(1) = 2 - 1 = 1 \). \( A^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} \). Comparing this with the given \( A^{-1} = \begin{bmatrix} 1 & -1 \\ x & 2 \end{bmatrix} \), we find that \( x = -1 \).

🎯 Exam Tip: The property \( AA^{-1} = I \) (Identity Matrix) is key. You can either calculate the inverse of A and compare, or multiply A by the given \( A^{-1} \) and equate it to I to solve for x.

Question 10. If \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), then matrix form is \[ \begin{bmatrix} ... & ... \\ ... & ... \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ... \\ ... \end{bmatrix} \]
Answer: \( \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \)
In simple words: A system of linear equations can be represented in matrix form as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. For the given equations, \( a_1 \) and \( b_1 \) are coefficients of the first equation, and \( a_2 \) and \( b_2 \) are coefficients of the second equation.

🎯 Exam Tip: Learn to convert systems of linear equations into matrix form (AX=B). The coefficient matrix A contains the coefficients of the variables, the variable matrix X contains the variables, and the constant matrix B contains the constants on the right side of the equations.

(III) State Whether Each Of The Following Is True Or False:

Question 1. Single element matrix is row as well as a column matrix.
Answer: True
In simple words: A single element matrix (e.g., [5]) has 1 row and 1 column. Since it has only one row, it can be considered a row matrix. Since it has only one column, it can also be considered a column matrix. Therefore, it is both.

🎯 Exam Tip: Understand the definitions of row and column matrices. A 1x1 matrix satisfies both definitions, making this statement true.

Question 2. Every scalar matrix is unit matrix.
Answer: False
In simple words: A scalar matrix has all its diagonal elements equal (e.g., \( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \)). A unit matrix (or identity matrix) is a scalar matrix where all diagonal elements are specifically 1 (e.g., \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)). So, not every scalar matrix is a unit matrix; only scalar matrices with 1s on the diagonal are unit matrices.

🎯 Exam Tip: Differentiate between scalar and unit (identity) matrices. While a unit matrix is a type of scalar matrix, the converse is not true. A scalar matrix can have any scalar value on its diagonal, not just 1.

Question 3. \( A = \begin{bmatrix} 4 & 5 \\ 6 & 1 \end{bmatrix} \) is non-singular matrix.
Answer: True
In simple words: A matrix is non-singular if its determinant is not equal to zero. For A, \( |A| = (4)(1) - (5)(6) = 4 - 30 = -26 \). Since \( -26 \neq 0 \), the matrix A is non-singular.

🎯 Exam Tip: To check if a matrix is non-singular, simply calculate its determinant. If the determinant is non-zero, the matrix is non-singular and therefore invertible.

Question 4. If A is symmetric, then \( A = -A^T \).
Answer: False
In simple words: A matrix A is symmetric if its transpose is equal to itself, i.e., \( A^T = A \). The statement \( A = -A^T \) is the definition of a skew-symmetric matrix. Therefore, the given statement is false.

🎯 Exam Tip: Clearly distinguish between symmetric (\( A^T = A \)) and skew-symmetric (\( A^T = -A \)) matrices. Understanding these definitions is fundamental for matrix properties and transformations.

Question 5. If AB and BA both exist, then AB = BA.
Answer: False
In simple words: Matrix multiplication is generally not commutative. Even if both products AB and BA exist (meaning the matrices are conformable for both multiplications, e.g., square matrices of the same order), their results are not necessarily equal. For them to be equal, the matrices must commute.

🎯 Exam Tip: Remember that matrix multiplication is NOT commutative in general. Only specific pairs of matrices satisfy AB = BA. This is a common point of confusion from scalar algebra.

Question 6. If A and B are square matrices of same order, then \( (A + B)^2 = A^2 + 2AB + B^2 \).
Answer: False
In simple words: When expanding \( (A+B)^2 \) for matrices, you get \( (A+B)(A+B) = A^2 + AB + BA + B^2 \). This is only equal to \( A^2 + 2AB + B^2 \) if \( AB = BA \) (i.e., A and B commute). Since matrix multiplication is generally not commutative, this identity is not always true.

🎯 Exam Tip: Be careful with matrix algebra; binomial expansion rules often do not directly apply unless commutativity (AB = BA) is guaranteed. Always expand \( (A+B)^2 \) as \( (A+B)(A+B) \).

Question 7. If A and B are conformable for the product AB, then \( (AB)^T = A^TB^T \).
Answer: False
In simple words: The correct property for the transpose of a product of matrices is \( (AB)^T = B^TA^T \). The order of multiplication for the transposes is reversed. This is a fundamental property of matrix transposition.

🎯 Exam Tip: Remember the reversal rule for the transpose of a product: \( (AB)^T = B^TA^T \). This is a common error point, so always ensure the order of the transposed matrices is reversed.

Question 8. Singleton matrix is only row matrix.
Answer: False
In simple words: A singleton matrix has only one element (e.g., [7]). It has 1 row and 1 column. It is both a row matrix (because it has 1 row) and a column matrix (because it has 1 column). So saying it's *only* a row matrix is incorrect.

🎯 Exam Tip: A 1x1 matrix is simultaneously a row matrix and a column matrix. Avoid limiting definitions to just one category if multiple apply.

Question 9. \( A = \begin{bmatrix} 2 & 1 \\ 10 & 5 \end{bmatrix} \) is invertible matrix.
Answer: False
In simple words: A matrix is invertible if and only if its determinant is non-zero (i.e., it is non-singular). For A, \( |A| = (2)(5) - (1)(10) = 10 - 10 = 0 \). Since the determinant is 0, the matrix A is singular and therefore not invertible.

🎯 Exam Tip: The invertibility of a matrix is directly linked to its determinant. A matrix is invertible if and only if its determinant is non-zero. Always check the determinant first to determine invertibility.

Question 10. \( A(\text{adj } A) = |A| I \), where I is the unit matrix.
Answer: True
In simple words: This is a fundamental property in matrix algebra. The product of a square matrix A and its adjoint (adj A) is always equal to the determinant of A multiplied by the identity (unit) matrix I.

🎯 Exam Tip: This identity, \( A(\text{adj } A) = |A|I \), is a cornerstone for understanding matrix inverses and their relationship with the adjoint matrix. It is essential for solving many matrix-related problems.

(IV) Solve The Following:

Question 1. Find k, if \( \begin{bmatrix} 7 & 3 \\ 5 & k \end{bmatrix} \) is a singular matrix.
Answer: Solution: Let \( A = \begin{bmatrix} 7 & 3 \\ 5 & k \end{bmatrix} \) Since, A is singular matrix, \( |A| = 0 \)
\( \therefore \begin{vmatrix} 7 & 3 \\ 5 & k \end{vmatrix} = 0 \)
\( \therefore 7k - 15 = 0 \)
\( \therefore k = \frac{15}{7} \)
In simple words: A matrix is singular if its determinant is zero. For a 2x2 matrix, the determinant is \( ad-bc \). By setting this expression equal to zero, we can solve for the unknown variable k.

🎯 Exam Tip: The concept of a singular matrix (determinant equals zero) is crucial. Accurately calculate the determinant and solve the resulting algebraic equation to find the unknown variable.

Question 2. Find x, y, z if \( \begin{bmatrix} 2 & x & 5 \\ 3 & 1 & z \\ y & 5 & 8 \end{bmatrix} \) is a symmetric matrix.
Answer: Solution: Let \( A = \begin{bmatrix} 2 & x & 5 \\ 3 & 1 & z \\ y & 5 & 8 \end{bmatrix} \)
Then \( A^T = \begin{bmatrix} 2 & 3 & y \\ x & 1 & 5 \\ 5 & z & 8 \end{bmatrix} \)
Since, A is symmetric matrix, \( A = A^T \)
\( \therefore \begin{bmatrix} 2 & x & 5 \\ 3 & 1 & z \\ y & 5 & 8 \end{bmatrix} = \begin{bmatrix} 2 & 3 & y \\ x & 1 & 5 \\ 5 & z & 8 \end{bmatrix} \)
By equality of matrices, x = 3, y = 5 and z = 5.
In simple words: A matrix is symmetric if it is equal to its transpose (\( A = A^T \)). This means that for every element \( a_{ij} \), it must be equal to its corresponding element \( a_{ji} \) in the transpose. By comparing the elements of A with those of \( A^T \), we can find the values of x, y, and z.

🎯 Exam Tip: The definition of a symmetric matrix (\( A = A^T \)) is key. This implies \( a_{ij} = a_{ji} \) for all i, j. Carefully equate corresponding elements of the matrix and its transpose to solve for unknowns.

Question 3. If \( A = \begin{bmatrix} 1 & 5 \\ 7 & 8 \\ 9 & 5 \end{bmatrix}, B = \begin{bmatrix} 2 & 4 \\ 1 & 5 \\ -8 & 6 \end{bmatrix}, C = \begin{bmatrix} -2 & 3 \\ 1 & -5 \\ 7 & 8 \end{bmatrix} \) then show that \( (A + B) + C = A + (B + C) \).
Answer: Solution: \[ A + B = \begin{bmatrix} 1 & 5 \\ 7 & 8 \\ 9 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 1 & 5 \\ -8 & 6 \end{bmatrix} \] \[ = \begin{bmatrix} 1+2 & 5+4 \\ 7+1 & 8+5 \\ 9+(-8) & 5+6 \end{bmatrix} \] \[ = \begin{bmatrix} 3 & 9 \\ 8 & 13 \\ 1 & 11 \end{bmatrix} \]
\( \therefore (A + B) + C = \begin{bmatrix} 3 & 9 \\ 8 & 13 \\ 1 & 11 \end{bmatrix} + \begin{bmatrix} -2 & 3 \\ 1 & -5 \\ 7 & 8 \end{bmatrix} \) \[ = \begin{bmatrix} 3+(-2) & 9+3 \\ 8+1 & 13+(-5) \\ 1+7 & 11+8 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 12 \\ 9 & 8 \\ 8 & 19 \end{bmatrix} \text{ ......(1)} \]
Also, \[ B + C = \begin{bmatrix} 2 & 4 \\ 1 & 5 \\ -8 & 6 \end{bmatrix} + \begin{bmatrix} -2 & 3 \\ 1 & -5 \\ 7 & 8 \end{bmatrix} \] \[ = \begin{bmatrix} 2+(-2) & 4+3 \\ 1+1 & 5+(-5) \\ -8+7 & 6+8 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 7 \\ 2 & 0 \\ -1 & 14 \end{bmatrix} \]
\( \therefore A + (B + C) = \begin{bmatrix} 1 & 5 \\ 7 & 8 \\ 9 & 5 \end{bmatrix} + \begin{bmatrix} 0 & 7 \\ 2 & 0 \\ -1 & 14 \end{bmatrix} \] \[ = \begin{bmatrix} 1+0 & 5+7 \\ 7+2 & 8+0 \\ 9+(-1) & 5+14 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 12 \\ 9 & 8 \\ 8 & 19 \end{bmatrix} \text{ ......(2)} \]
From (1) and (2), \( (A + B) + C = A + (B + C) \).
In simple words: This problem demonstrates the associative property of matrix addition, which states that \( (A+B)+C = A+(B+C) \). You need to first calculate A+B and then add C to the result. Separately, calculate B+C and then add A to that result. If both final matrices are identical, the property is shown to hold.

🎯 Exam Tip: Matrix addition is associative. This means the grouping of matrices during addition does not affect the final sum. Perform step-by-step matrix additions accurately and ensure all corresponding elements are correctly summed.

Question 4. If \( A = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix}, B = \begin{bmatrix} 1 & 7 \\ -3 & 0 \end{bmatrix} \) find the matrix \( A - 4B + 7I \), where I is the unit matrix of order 2.
Answer: Solution: \[ A - 4B + 7I = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} - 4 \begin{bmatrix} 1 & 7 \\ -3 & 0 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} - \begin{bmatrix} 4 \times 1 & 4 \times 7 \\ 4 \times (-3) & 4 \times 0 \end{bmatrix} + \begin{bmatrix} 7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} - \begin{bmatrix} 4 & 28 \\ -12 & 0 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \] \[ = \begin{bmatrix} 2-4+7 & 5-28+0 \\ 3-(-12)+0 & 7-0+7 \end{bmatrix} \] \[ = \begin{bmatrix} 5 & -23 \\ 15 & 14 \end{bmatrix} \]
In simple words: This problem involves scalar multiplication of matrices and matrix addition/subtraction. First, multiply matrix B by 4 and identity matrix I by 7. Then, perform the subtraction of 4B from A and finally add 7I to the result, by performing element-wise operations.

🎯 Exam Tip: Remember that scalar multiplication involves multiplying every element of the matrix by the scalar. Matrix addition/subtraction is performed element-wise. The identity matrix (I) must be of the same order as A and B for addition/subtraction to be conformable.

Question 5. If \( A = \begin{bmatrix} 2 & -3 \\ 3 & -2 \\ -1 & 4 \end{bmatrix}, B = \begin{bmatrix} -3 & 4 & 1 \\ 2 & -1 & -3 \end{bmatrix} \) verify
(i) \( (A + 2B^T)^T = A^T + 2B \)
(ii) \( (3A - 5B^T)^T = 3A^T - 5B \)
Answer: Solution: Given matrices are: \[ A = \begin{bmatrix} 2 & -3 \\ 3 & -2 \\ -1 & 4 \end{bmatrix}, B = \begin{bmatrix} -3 & 4 & 1 \\ 2 & -1 & -3 \end{bmatrix} \]
First, find the transposes: \[ A^T = \begin{bmatrix} 2 & 3 & -1 \\ -3 & -2 & 4 \end{bmatrix} \] \[ B^T = \begin{bmatrix} -3 & 2 \\ 4 & -1 \\ 1 & -3 \end{bmatrix} \]
(i) \( (A + 2B^T)^T = A^T + 2B \)
LHS: \[ A + 2B^T = \begin{bmatrix} 2 & -3 \\ 3 & -2 \\ -1 & 4 \end{bmatrix} + 2 \begin{bmatrix} -3 & 2 \\ 4 & -1 \\ 1 & -3 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & -3 \\ 3 & -2 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} -6 & 4 \\ 8 & -2 \\ 2 & -6 \end{bmatrix} \] \[ = \begin{bmatrix} 2-6 & -3+4 \\ 3+8 & -2-2 \\ -1+2 & 4-6 \end{bmatrix} \] \[ = \begin{bmatrix} -4 & 1 \\ 11 & -4 \\ 1 & -2 \end{bmatrix} \]
\( \therefore (A + 2B^T)^T = \begin{bmatrix} -4 & 11 & 1 \\ 1 & -4 & -2 \end{bmatrix} \text{ ......(1)} \)
RHS: \[ A^T + 2B = \begin{bmatrix} 2 & 3 & -1 \\ -3 & -2 & 4 \end{bmatrix} + 2 \begin{bmatrix} -3 & 4 & 1 \\ 2 & -1 & -3 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & 3 & -1 \\ -3 & -2 & 4 \end{bmatrix} + \begin{bmatrix} -6 & 8 & 2 \\ 4 & -2 & -6 \end{bmatrix} \] \[ = \begin{bmatrix} 2-6 & 3+8 & -1+2 \\ -3+4 & -2-2 & 4-6 \end{bmatrix} \] \[ = \begin{bmatrix} -4 & 11 & 1 \\ 1 & -4 & -2 \end{bmatrix} \text{ ......(2)} \]
From (1) and (2), \( (A + 2B^T)^T = A^T + 2B \).
(ii) \( (3A - 5B^T)^T = 3A^T - 5B \)
LHS: \[ 3A - 5B^T = 3 \begin{bmatrix} 2 & -3 \\ 3 & -2 \\ -1 & 4 \end{bmatrix} - 5 \begin{bmatrix} -3 & 2 \\ 4 & -1 \\ 1 & -3 \end{bmatrix} \] \[ = \begin{bmatrix} 6 & -9 \\ 9 & -6 \\ -3 & 12 \end{bmatrix} - \begin{bmatrix} -15 & 10 \\ 20 & -5 \\ 5 & -15 \end{bmatrix} \] \[ = \begin{bmatrix} 6-(-15) & -9-10 \\ 9-20 & -6-(-5) \\ -3-5 & 12-(-15) \end{bmatrix} \] \[ = \begin{bmatrix} 21 & -19 \\ -11 & -1 \\ -8 & 27 \end{bmatrix} \]
\( \therefore (3A - 5B^T)^T = \begin{bmatrix} 21 & -11 & -8 \\ -19 & -1 & 27 \end{bmatrix} \text{ ......(1)} \)
RHS: \[ 3A^T - 5B = 3 \begin{bmatrix} 2 & 3 & -1 \\ -3 & -2 & 4 \end{bmatrix} - 5 \begin{bmatrix} -3 & 4 & 1 \\ 2 & -1 & -3 \end{bmatrix} \] \[ = \begin{bmatrix} 6 & 9 & -3 \\ -9 & -6 & 12 \end{bmatrix} - \begin{bmatrix} -15 & 20 & 5 \\ 10 & -5 & -15 \end{bmatrix} \] \[ = \begin{bmatrix} 6-(-15) & 9-20 & -3-5 \\ -9-10 & -6-(-5) & 12-(-15) \end{bmatrix} \] \[ = \begin{bmatrix} 21 & -11 & -8 \\ -19 & -1 & 27 \end{bmatrix} \text{ ......(2)} \]
From (1) and (2), \( (3A - 5B^T)^T = 3A^T - 5B \).
In simple words: This problem verifies two properties related to matrix transpose, scalar multiplication, and addition/subtraction. First, calculate the transposes of A and B. Then, for each part, calculate the Left Hand Side (LHS) by performing operations inside the bracket first, then transposing the result. Separately, calculate the Right Hand Side (RHS) by performing scalar multiplication and then addition/subtraction. Compare the LHS and RHS to verify the identity. Remember \( (A \pm B)^T = A^T \pm B^T \) and \( (kA)^T = kA^T \).

🎯 Exam Tip: Be meticulous with matrix transpose properties: \( (A+B)^T = A^T+B^T \) and \( (kA)^T = kA^T \). Remember that the order of operations matters. Calculate scalar multiples and additions/subtractions element-wise before transposing the final result. Verify both sides independently.

Question 6. If \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0 \end{bmatrix} \) then show that AB and BA are both singular matrices.

Solution:

\( AB = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 1-6-6 & -1+4+3 & 1-2+0 \\ 2-12-12 & -2+8+6 & 2-4+0 \\ 1-6-6 & -1+4+3 & 1-2+0 \end{bmatrix} \)
\( = \begin{bmatrix} -11 & 6 & -1 \\ -22 & 12 & -2 \\ -11 & 6 & -1 \end{bmatrix} \)
\( \implies |AB| = \begin{vmatrix} -11 & 6 & -1 \\ -22 & 12 & -2 \\ -11 & 6 & -1 \end{vmatrix} \)
By taking -6 common from \( C_2 \), we get
\( \implies |AB| = -6 \begin{vmatrix} -11 & -1 & -1 \\ -22 & -2 & -2 \\ -11 & -1 & -1 \end{vmatrix} \)
\( = -6 \times 0 \)
\( = 0 \)
\( \implies \) AB is a singular matrix.
Also, \( BA = \begin{bmatrix} 1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 1-2+1 & 2-4+2 & 3-6+3 \\ -3+4-1 & -6+8-2 & -9+12-3 \\ -2+2+0 & -4+4+0 & -6+6+0 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
\( \implies |BA| = 0 \)
\( \implies \) BA is also a singular matrix.
Hence, AB and BA are both singular matrices.
In simple words: This problem asks us to multiply two matrices, A and B, in both orders (AB and BA), and then show that the determinant of the resulting matrices is zero, which means they are singular. Singular matrices are those that do not have an inverse.

🎯 Exam Tip: Remember that a square matrix is singular if and only if its determinant is zero. For matrix multiplication, carefully follow the row-by-column multiplication rule to avoid errors.

Question 7. If \( A = \begin{bmatrix} 3 & 1 \\ 1 & 5 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 2 \\ 5 & -2 \end{bmatrix} \), verify \( |AB| = |A||B| \).

Solution:

\( AB = \begin{bmatrix} 3 & 1 \\ 1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 5 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 3+5 & 6-2 \\ 1+25 & 2-10 \end{bmatrix} \)
\( = \begin{bmatrix} 8 & 4 \\ 26 & -8 \end{bmatrix} \)
\( \implies |AB| = \begin{vmatrix} 8 & 4 \\ 26 & -8 \end{vmatrix} \)
\( = 8(-8) - 4(26) \)
\( = -64 - 104 = -168 \) ... (1)
\( |A| = \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} \)
\( = 3(5) - 1(1) = 15-1 = 14 \)
\( |B| = \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} \)
\( = 1(-2) - 2(5) = -2-10 = -12 \)
\( \implies |A||B| = 14(-12) = -168 \) ... (2)
From (1) and (2), \( |AB| = |A||B| \).
In simple words: This problem verifies a fundamental property of determinants for matrices, stating that the determinant of a product of matrices is equal to the product of their determinants. We calculate AB, then find its determinant. Separately, we find the determinants of A and B, multiply them, and confirm the results are the same.

🎯 Exam Tip: This property \( |AB| = |A||B| \) is crucial for matrix theory and often appears in exams. Ensure accurate calculation of matrix products and determinants.

Question 8. If \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \), then show that \( A^2 - 4A + 3I = 0 \).

Solution:

\( A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \)
\( \implies A^2 - 4A + 3I = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} - 4 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} + 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} - \begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 5-8+3 & -4-(-4)+0 \\ -4-(-4)+0 & 5-8+3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \implies A^2 - 4A + 3I = 0 \).
In simple words: We need to prove that a given matrix A satisfies a specific polynomial equation. We calculate \( A^2 \) by multiplying A by itself, then perform scalar multiplication for \( 4A \) and \( 3I \) (where I is the identity matrix), and finally add/subtract the resulting matrices to show the final result is a zero matrix.

🎯 Exam Tip: Remember that \( I \) represents the identity matrix of the same order as \( A \). Accurate matrix multiplication, scalar multiplication, and addition/subtraction are essential for solving such problems correctly.

Question 9. If \( A = \begin{bmatrix} -3 & 2 \\ 2 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & a \\ b & 0 \end{bmatrix} \) and \( (A+B)(A-B) = A^2 - B^2 \), find a and b.

Solution:

\( (A+B)(A-B) = A^2 - B^2 \)
\( \implies A^2 - AB + BA - B^2 = A^2 - B^2 \)
\( \implies -AB + BA = 0 \)
\( \implies AB = BA \)
Now, \( AB = \begin{bmatrix} -3 & 2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & a \\ b & 0 \end{bmatrix} = \begin{bmatrix} -3+2b & -3a+0 \\ 2+4b & 2a+0 \end{bmatrix} \)
And \( BA = \begin{bmatrix} 1 & a \\ b & 0 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -3+2a & 2+4a \\ -3b+0 & 2b+0 \end{bmatrix} \)
By equality of matrices,
\( -3+2b = -3+2a \) ........(1)
\( -3a = 2+4a \) ........(2)
\( 2+4b = -3b \) ........(3)
\( 2a = 2b \) ........(4)
From (2), \( -3a = 2+4a \)
\( \implies -7a = 2 \)
\( \implies a = -\frac{2}{7} \)
From (3), \( 2+4b = -3b \)
\( \implies 7b = -2 \)
\( \implies b = -\frac{2}{7} \)
These values of a and b also satisfy equations (1) and (4).
Hence, \( a = -\frac{2}{7} \) and \( b = -\frac{2}{7} \).
In simple words: The problem provides a condition for matrix products, \( (A+B)(A-B) = A^2 - B^2 \). This implies that AB and BA are equal. By multiplying the matrices A and B in both orders and setting the resulting matrices equal to each other, we can compare their elements to find the values of a and b.

🎯 Exam Tip: Remember that \( (A+B)(A-B) = A^2 - B^2 \) only holds true for matrices if AB = BA. Carefully perform matrix multiplication and element-wise comparison for accuracy.

Question 10. If \( A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \), then find \( A^3 \).

Solution:

\( A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 1-2 & 2+6 \\ -1-3 & -2+9 \end{bmatrix} \)
\( = \begin{bmatrix} -1 & 8 \\ -4 & 7 \end{bmatrix} \)
\( \implies A^3 = A^2 \cdot A = \begin{bmatrix} -1 & 8 \\ -4 & 7 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} -1-8 & -2+24 \\ -4-7 & -8+21 \end{bmatrix} \)
\( = \begin{bmatrix} -9 & 22 \\ -11 & 13 \end{bmatrix} \)
In simple words: To find \( A^3 \), we first calculate \( A^2 \) by multiplying matrix A by itself. Then, we multiply the resulting \( A^2 \) matrix by A again. Remember that matrix multiplication is not commutative.

🎯 Exam Tip: When calculating powers of matrices, break it down into successive multiplications (e.g., \( A^3 = A \cdot A \cdot A \)). Precision in each multiplication step is vital for the final answer.

Question 11. Find x, y, z if \( \begin{bmatrix} 0 & 1 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & -2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix} \)

Solution:

\( \begin{bmatrix} 0 & 1 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & -2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix} \)
\( \implies \begin{bmatrix} 0(2)+1(3)-1(1) \\ 5(2)+1(3)+0(1) \\ 1(2)+1(3)-1(1) \end{bmatrix} = \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix} \)
\( \implies \begin{bmatrix} 0+3-1 \\ 10+3+0 \\ 2+3-1 \end{bmatrix} = \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix} \)
\( \implies \begin{bmatrix} 2 \\ 13 \\ 4 \end{bmatrix} = \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix} \)
By equality of matrices, we get
\( x-1 = 2 \implies x = 3 \)
\( y+1 = 13 \implies y = 12 \)
\( 2z = 4 \implies z = 2 \)
In simple words: We are given a matrix multiplication on the left side and a column matrix with unknown variables on the right. First, we perform the matrix multiplication. Then, by equating the corresponding elements of the resulting matrices, we form equations to solve for x, y, and z.

🎯 Exam Tip: Be meticulous with matrix multiplication, ensuring each element in the resulting matrix is correctly calculated. After multiplication, equate corresponding elements carefully to set up the correct linear equations.

Question 12. If \( A = \begin{bmatrix} 2 & -4 \\ 3 & -2 \\ 0 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 1 & 0 \end{bmatrix} \) then show that \( (AB)^T = B^T A^T \).

Solution:

\( AB = \begin{bmatrix} 2 & -4 \\ 3 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ -2 & 1 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 2+8 & -2-4 & 4+0 \\ 3+4 & -3-2 & 6-0 \\ 0-2 & 0+1 & 0+0 \end{bmatrix} \)
\( = \begin{bmatrix} 10 & -6 & 4 \\ 7 & -5 & 6 \\ -2 & 1 & 0 \end{bmatrix} \)
\( \implies (AB)^T = \begin{bmatrix} 10 & 7 & -2 \\ -6 & -5 & 1 \\ 4 & 6 & 0 \end{bmatrix} \) ... (i)
Now, \( A^T = \begin{bmatrix} 2 & 3 & 0 \\ -4 & -2 & 1 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 1 & -2 \\ -1 & 1 \\ 2 & 0 \end{bmatrix} \)
\( \implies B^T A^T = \begin{bmatrix} 1 & -2 \\ -1 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 3 & 0 \\ -4 & -2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2+8 & 3+4 & 0-2 \\ -2-4 & -3-2 & 0+1 \\ 4+0 & 6+0 & 0+0 \end{bmatrix} \)
\( = \begin{bmatrix} 10 & 7 & -2 \\ -6 & -5 & 1 \\ 4 & 6 & 0 \end{bmatrix} \) ... (ii)
From (i) and (ii), we get
\( (AB)^T = B^T A^T \).
In simple words: This problem asks us to verify the transpose property for matrix products. First, we calculate the product AB and then find its transpose. Separately, we find the transposes of A and B, then multiply them in reverse order (\( B^T A^T \)), and confirm that both results are identical.

🎯 Exam Tip: Remember that \( (AB)^T = B^T A^T \), not \( A^T B^T \). Careful calculation of matrix products and transposes is key to proving this identity.

Question 13. If \( A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix} \), then reduce it to unit matrix by row transformation.

Solution:

\( A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix} \)
\( = 1(1-0) - 0(2-0) + 0(6-3) \)
\( = 1-0+0 \)
\( = 1 \neq 0 \)
\( \implies \) A is non-singular matrix.
Hence, row transformations are possible.
Now, \( A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix} \)
Applying \( R_2 \to R_2-2R_1 \) and \( R_3 \to R_3-3R_1 \), we get
\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 3 & 1 \end{bmatrix} \)
Applying \( R_3 \to R_3-3R_2 \), we get
\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
In simple words: We start with matrix A and perform elementary row operations to transform it into an identity matrix. First, we make the elements below the leading 1 in the first column zero. Then, we make the element below the leading 1 in the second column zero. This process transforms A into an identity matrix.

🎯 Exam Tip: Remember the goal of row transformations is to achieve the identity matrix. Always target zeros below the leading 1s, and then ones on the diagonal. Performing operations simultaneously on multiple rows can save time, but ensure accuracy.

Question 14. Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rs.) of these crops by both the farmers for the month of April and May 2016 is given below:

April 2016 (In Rs.)

May 2016 (In Rs.)

Find (i) the total sale in rupees for two months of each farmer for each crop. (ii) the increase in sales from April to May for every crop of each farmer.

Solution:

The given information can be written in matrix form as:

April 2016 (in Rs.)

\( A = \begin{bmatrix} \text{Rice} & \text{Wheat} & \text{Groundnut} \\ 15000 & 13000 & 12000 \\ 18000 & 15000 & 8000 \end{bmatrix} \begin{matrix} \text{Shantaram} \\ \text{Kantaram} \end{matrix} \)

May 2016 (in Rs.)

\( B = \begin{bmatrix} \text{Rice} & \text{Wheat} & \text{Groundnut} \\ 18000 & 15000 & 12000 \\ 21000 & 16500 & 16000 \end{bmatrix} \begin{matrix} \text{Shantaram} \\ \text{Kantaram} \end{matrix} \)

(i) The total sale in Rs. for two months of each farmer for each crop can be obtained by the addition A + B.
Now, \( A+B = \begin{bmatrix} 15000 & 13000 & 12000 \\ 18000 & 15000 & 8000 \end{bmatrix} + \begin{bmatrix} 18000 & 15000 & 12000 \\ 21000 & 16500 & 16000 \end{bmatrix} \)
\( = \begin{bmatrix} 15000+18000 & 13000+15000 & 12000+12000 \\ 18000+21000 & 15000+16500 & 8000+16000 \end{bmatrix} \)
\( = \begin{bmatrix} 33000 & 28000 & 24000 \\ 39000 & 31500 & 24000 \end{bmatrix} \)
\( \implies \) total sale in Rs. for two months of each farmer for each crop is given by
\( \begin{bmatrix} \text{Rice} & \text{Wheat} & \text{Groundnut} \\ 33000 & 28000 & 24000 \\ 39000 & 31500 & 24000 \end{bmatrix} \begin{matrix} \text{Shantaram} \\ \text{Kantaram} \end{matrix} \)
Hence, the total sale for Shantaram are Rs. 33000 for Rice, Rs. 28000 for Wheat, Rs. 24000 for Groundnut, and for Kantaram are Rs. 39000 for Rice, Rs. 31500 for Wheat, Rs. 24000 for Groundnut.
(ii) The increase in sales from April to May for every crop of each farmer can be obtained by the subtraction of A from B.
Now, \( B-A = \begin{bmatrix} 18000 & 15000 & 12000 \\ 21000 & 16500 & 16000 \end{bmatrix} - \begin{bmatrix} 15000 & 13000 & 12000 \\ 18000 & 15000 & 8000 \end{bmatrix} \)
\( = \begin{bmatrix} 18000-15000 & 15000-13000 & 12000-12000 \\ 21000-18000 & 16500-15000 & 16000-8000 \end{bmatrix} \)
\( = \begin{bmatrix} 3000 & 2000 & 0 \\ 3000 & 1500 & 8000 \end{bmatrix} \)
\( \implies \) the increase in sale from April to May is given by
\( \begin{bmatrix} \text{Rice} & \text{Wheat} & \text{Groundnut} \\ 3000 & 2000 & 0 \\ 3000 & 1500 & 8000 \end{bmatrix} \begin{matrix} \text{Shantaram} \\ \text{Kantaram} \end{matrix} \)
Hence, the increase in sales from April to May of Shantam is Rs. 3000 in Rice, Rs. 2000 in Wheat, nothing in Groundnut and of Kantaram are Rs. 3000 in Rice, Rs. 1500 in Wheat, Rs. 8000 in Groundnut.
In simple words: We represent the monthly sales data for two farmers across three crops as matrices. To find the total sales over two months, we add the corresponding matrices. To find the increase in sales from April to May, we subtract the April sales matrix from the May sales matrix. This provides a clear overview of their agricultural business performance.

🎯 Exam Tip: When dealing with real-world data in matrix form, pay close attention to the units (e.g., Rs.) and interpret the results correctly in context. Element-wise addition and subtraction are straightforward, but ensure all numbers are handled precisely.

Question 15. Check whether following matrices are invertible or not:
(i) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)

Solution:

(i) Let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} \)
\( = 1-0 \)
\( = 1 \neq 0 \)
\( \implies \) A is a non-singular matrix.
Hence, \( A^{-1} \) exists.
(ii) Let \( A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} \)
\( = 1-1 \)
\( = 0 \)
\( \implies \) A is a singular matrix.
Hence, \( A^{-1} \) does not exist.
(iii) Let \( A = \begin{bmatrix} 3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{vmatrix} \)
\( = 3(5-0) - 4(5-0) + 3(4-1) \)
\( = 15-20+9 \)
\( = 4 \neq 0 \)
\( \implies \) A is a non-singular matrix.
Hence, \( A^{-1} \) exists.
(iv) Let \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 2 & 4 & 6 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 2 & 4 & 6 \end{vmatrix} \)
\( = 1(24-20) - 2(12-10) + 3(8-8) \)
\( = 4-4+0 \)
\( = 0 \)
\( \implies \) A is a singular matrix.
Hence, \( A^{-1} \) does not exist.
In simple words: To check if a matrix is invertible, we calculate its determinant. If the determinant is non-zero, the matrix is non-singular and its inverse exists. If the determinant is zero, the matrix is singular and its inverse does not exist.

🎯 Exam Tip: The invertibility of a matrix is directly tied to its determinant. Practice calculating determinants for 2x2 and 3x3 matrices accurately to quickly determine invertibility.

Question 16. Find inverse of the following matrices (if they exist) by elementary transformation:
(i) \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)

Solution:

Let \( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 1(3) - (-1)(2) = 3 - (-2) = 5 \neq 0 \)
\( \implies A^{-1} \) exists.
We write, \( AA^{-1} = I \)
\( \implies \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
By \( R_2 \to R_2-2R_1 \), we get
\( \begin{bmatrix} 1 & -1 \\ 0 & 5 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \)
By \( R_2 \to \frac{1}{5}R_2 \), we get
\( \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 \\ -\frac{2}{5} & \frac{1}{5} \end{bmatrix} \)
By \( R_1 \to R_1+R_2 \), we get
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1-\frac{2}{5} & 0+\frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{bmatrix} \)
\( \implies A^{-1} = \begin{bmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{bmatrix} \)
(ii) Let \( A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 2 & 1 \\ 7 & 4 \end{vmatrix} = 2(4) - 1(7) = 8-7 = 1 \neq 0 \)
\( \implies A^{-1} \) exist.
We write \( A A^{-1} = I \)
\( \implies \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
By \( C_1 \to C_1-C_2 \), we get
\( \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \)
By \( C_2 \to C_2-C_1 \), we get
\( \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} \)
By \( C_1 \to C_1-3C_2 \), we get
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix} \)
\( \implies A^{-1} = \begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix} \)
(iii) Let \( A = \begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix} \)
Then \( |A| = \begin{vmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{vmatrix} \)
\( = 2(4+6) + 3(4-9) + 3(-4-6) \)
\( = 2(10) + 3(-5) + 3(-10) \)
\( = 20-15-30 = -25 \neq 0 \)
\( \implies A^{-1} \) exists.
We write \( A A^{-1} = I \)
\( \implies \begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( C_1 \leftrightarrow C_3 \), we get
\( \begin{bmatrix} 3 & -3 & 2 \\ 3 & 2 & 2 \\ 2 & -2 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
By \( C_1 \to C_1-C_3 \), we get
\( \begin{bmatrix} 1 & -3 & 2 \\ 1 & 2 & 2 \\ -1 & -2 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
By \( C_2 \to C_2+3C_1 \) and \( C_3 \to C_3-2C_1 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 1 & 5 & 0 \\ -1 & -5 & 5 \end{bmatrix} A^{-1} = \begin{bmatrix} -1 & -3 & 3 \\ 0 & 1 & 0 \\ 1 & 3 & -2 \end{bmatrix} \)
By \( C_2 \to \frac{1}{5} C_2 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 5 \end{bmatrix} A^{-1} = \begin{bmatrix} -1 & -\frac{3}{5} & 3 \\ 0 & \frac{1}{5} & 0 \\ 1 & \frac{3}{5} & -2 \end{bmatrix} \)
By \( C_1 \to C_1-C_2 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 5 \end{bmatrix} A^{-1} = \begin{bmatrix} -1+\frac{3}{5} & -\frac{3}{5} & 3 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ 1-\frac{3}{5} & \frac{3}{5} & -2 \end{bmatrix} = \begin{bmatrix} -\frac{2}{5} & -\frac{3}{5} & 3 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{3}{5} & -2 \end{bmatrix} \)
By \( C_3 \to \frac{1}{5} C_3 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} -\frac{2}{5} & -\frac{3}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{3}{5} & -\frac{2}{5} \end{bmatrix} \)
By \( C_1 \to C_1+C_3 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} -\frac{2}{5}+\frac{3}{5} & -\frac{3}{5} & \frac{3}{5} \\ -\frac{1}{5}+0 & \frac{1}{5} & 0 \\ \frac{2}{5}-\frac{2}{5} & \frac{3}{5} & -\frac{2}{5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & -\frac{3}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & \frac{3}{5} & -\frac{2}{5} \end{bmatrix} \)
\( \implies A^{-1} = \frac{1}{5} \begin{bmatrix} 1 & -3 & 3 \\ -1 & 1 & 0 \\ 0 & 3 & -2 \end{bmatrix} \)
(iv) Let \( A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \)
\( \implies |A| = \begin{vmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{vmatrix} \)
\( = 2(3-0) - 0(15-0) - 1(5-0) \)
\( = 6-0-5 \)
\( = 1 \neq 0 \)
\( \implies A^{-1} \) exists.
We write \( AA^{-1} = I \)
\( \implies \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( R_1 \leftrightarrow R_2 \), we get
\( \begin{bmatrix} 5 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( R_1 \to R_1-2R_2 \), we get
\( \begin{bmatrix} 1 & 1 & 2 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( R_2 \to R_2-2R_1 \), we get
\( \begin{bmatrix} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( R_2 \to R_2+3R_3 \), we get
\( \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 4 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -2 & 1 & 0 \\ 5 & -2 & 3 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( R_1 \to R_1-R_2 \) and \( R_3 \to R_3-R_2 \), we get
\( \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & -1 \end{bmatrix} A^{-1} = \begin{bmatrix} -7 & 3 & -3 \\ 5 & -2 & 3 \\ -5 & 2 & -2 \end{bmatrix} \)
By \( R_3 \to (-1)R_3 \), we get
\( \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} -7 & 3 & -3 \\ 5 & -2 & 3 \\ 5 & -2 & 2 \end{bmatrix} \)
By \( R_1 \to R_1+2R_3 \) and \( R_2 \to R_2-4R_3 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} -7+10 & 3-4 & -3+4 \\ 5-20 & -2+8 & 3-8 \\ 5 & -2 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \)
\( \implies A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \)
In simple words: To find the inverse of a matrix using elementary transformations, we set up an augmented matrix \([A|I]\) and perform row (or column) operations to transform the left side (A) into an identity matrix (I). The same operations applied to the right side (I) will transform it into the inverse matrix (\(A^{-1}\)). First, verify that the determinant is non-zero to ensure an inverse exists.

🎯 Exam Tip: Elementary transformations require careful, step-by-step application. Focus on making elements 1 on the main diagonal and 0s elsewhere in a methodical manner (e.g., column by column or row by row). One mistake can propagate through the entire calculation.

Question 16. Find inverse of the following matrices (if they exist) by elementary transformation: (iv) \[ \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \] Solution: Let \[ A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \] \( |A| = \begin{vmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{vmatrix} \) \( = 2(3-0) - 0 - 1(5-0) \) \( = 6-0-5 \) \( = 1 \neq 0 \)
Therefore, \( A^{-1} \) exists. We write \( AA^{-1}=I \)
Therefore, \[ \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] By \( R_1 \leftrightarrow R_2 \), we get \[ \begin{bmatrix} 5 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] By \( R_1 - 2R_2 \), we get \[ \begin{bmatrix} 1 & 1 & 2 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] By \( R_2 - 2R_1 \), we get \[ \begin{bmatrix} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] By \( R_2 + 3R_3 \), we get \[ \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 4 \\ 0 & 1 & 3 \end{bmatrix} A^{-1} = \begin{bmatrix} -2 & 1 & 0 \\ 5 & -2 & 3 \\ 0 & 0 & 1 \end{bmatrix} \] By \( R_1 - R_2 \) and \( R_3 - R_2 \), we get \[ \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & -1 \end{bmatrix} A^{-1} = \begin{bmatrix} -7 & 3 & -3 \\ 5 & -2 & 3 \\ -5 & 2 & -2 \end{bmatrix} \] By \( (-1)R_3 \), we get \[ \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} -7 & 3 & -3 \\ 5 & -2 & 3 \\ 5 & -2 & 2 \end{bmatrix} \] By \( R_1 + 2R_3 \) and \( R_2 - 4R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \]
Therefore, \[ A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \] In simple words: The inverse of the given matrix is found by applying elementary row transformations to transform the original matrix into an identity matrix while simultaneously performing the same transformations on an identity matrix to get the inverse.

🎯 Exam Tip: Mastering elementary row transformations is crucial for finding matrix inverses. Ensure precise calculations at each step to avoid errors.

Question 17. Find the inverse of \[ \begin{bmatrix} 3 & 1 & 5 \\ 2 & 7 & 8 \\ 1 & 2 & 5 \end{bmatrix} \] by adjoint method. Solution: Let \[ A = \begin{bmatrix} 3 & 1 & 5 \\ 2 & 7 & 8 \\ 1 & 2 & 5 \end{bmatrix} \] Then \( |A| = \begin{vmatrix} 3 & 1 & 5 \\ 2 & 7 & 8 \\ 1 & 2 & 5 \end{vmatrix} \) \( = 3(35-16) - 1(10-8) + 5(4-7) \) \( = 3(19) - 1(2) + 5(-3) \) \( = 57-2-15 = 40 \neq 0 \)
Therefore, \( A^{-1} \) exists. First, we have to find the cofactor matrix \( = [A_{ij}]_{3 \times 3} \) where \( A_{ij} = (-1)^{i+j}M_{ij} \) Now, \( A_{11} = (-1)^{1+1}M_{11} = \begin{vmatrix} 7 & 8 \\ 2 & 5 \end{vmatrix} = 35-16 = 19 \) \( A_{12} = (-1)^{1+2}M_{12} = - \begin{vmatrix} 2 & 8 \\ 1 & 5 \end{vmatrix} = -(10-8) = -2 \) \( A_{13} = (-1)^{1+3}M_{13} = \begin{vmatrix} 2 & 7 \\ 1 & 2 \end{vmatrix} = 4-7 = -3 \) \( A_{21} = (-1)^{2+1}M_{21} = - \begin{vmatrix} 1 & 5 \\ 2 & 5 \end{vmatrix} = -(5-10) = 5 \) \( A_{22} = (-1)^{2+2}M_{22} = \begin{vmatrix} 3 & 5 \\ 1 & 5 \end{vmatrix} = 15-5 = 10 \) \( A_{23} = (-1)^{2+3}M_{23} = - \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = -(6-1) = -5 \) \( A_{31} = (-1)^{3+1}M_{31} = \begin{vmatrix} 1 & 5 \\ 7 & 8 \end{vmatrix} = 8-35 = -27 \) \( A_{32} = (-1)^{3+2}M_{32} = - \begin{vmatrix} 3 & 5 \\ 2 & 8 \end{vmatrix} = -(24-10) = -14 \) \( A_{33} = (-1)^{3+3}M_{33} = \begin{vmatrix} 3 & 1 \\ 2 & 7 \end{vmatrix} = 21-2 = 19 \)
Therefore, the cofactor matrix is \[ \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} 19 & -2 & -3 \\ 5 & 10 & -5 \\ -27 & -14 & 19 \end{bmatrix} \]
Therefore, adj \( A = \begin{bmatrix} 19 & 5 & -27 \\ -2 & 10 & -14 \\ -3 & -5 & 19 \end{bmatrix} \)
Therefore, \( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
Therefore, \( A^{-1} = \frac{1}{40} \begin{bmatrix} 19 & 5 & -27 \\ -2 & 10 & -14 \\ -3 & -5 & 19 \end{bmatrix} \) In simple words: To find the inverse of a matrix using the adjoint method, first calculate the determinant of the matrix. Then, find the cofactor matrix and its transpose (which is the adjoint). Finally, divide the adjoint matrix by the determinant.

🎯 Exam Tip: The adjoint method for inverse calculation is prone to sign errors. Double-check each cofactor calculation, especially the \((-1)^{i+j}\) term, and ensure the determinant is calculated correctly.

Question 18. Solve the following equations by method of inversion: (i) \( 4x - 3y - 2 = 0 \), \( 3x - 4y + 6 = 0 \) Solution: The given equations are \( 4x - 3y = 2 \) \( 3x - 4y = -6 \) These equations can be written in matrix form as: \[ \begin{bmatrix} 4 & -3 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ -6 \end{bmatrix} \] This is of the form \( AX = B \), where \[ A = \begin{bmatrix} 4 & -3 \\ 3 & -4 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 \\ -6 \end{bmatrix} \] Let us find \( A^{-1} \). \( |A| = \begin{vmatrix} 4 & -3 \\ 3 & -4 \end{vmatrix} \) \( = -16 - (-9) = -16 + 9 = -7 \neq 0 \)
Therefore, \( A^{-1} \) exists. We write \( AA^{-1}=I \)
Therefore, \[ \begin{bmatrix} 4 & -3 \\ 3 & -4 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] By \( R_1 - R_2 \), we get \[ \begin{bmatrix} 1 & 1 \\ 3 & -4 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \] By \( R_2 - 3R_1 \), we get \[ \begin{bmatrix} 1 & 1 \\ 0 & -7 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix} \] By \( (-\frac{1}{7})R_2 \), we get \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & -1 \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix} \] By \( R_1 - R_2 \), we get \[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1-\frac{3}{7} & -1-(-\frac{4}{7}) \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix} = \begin{bmatrix} \frac{4}{7} & -\frac{3}{7} \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix} \]
Therefore, \[ A^{-1} = \begin{bmatrix} \frac{4}{7} & -\frac{3}{7} \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix} \] Now, premultiply \( AX = B \) by \( A^{-1} \), we get \( A^{-1}(AX) = A^{-1}B \)
\( \implies (A^{-1}A)X = A^{-1}B \)
\( \implies IX = A^{-1}B \)
Therefore, \( X = A^{-1}B \)
Therefore, \[ X = \begin{bmatrix} \frac{4}{7} & -\frac{3}{7} \\ \frac{3}{7} & -\frac{4}{7} \end{bmatrix} \begin{bmatrix} 2 \\ -6 \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{4}{7}(2) + (-\frac{3}{7})(-6) \\ \frac{3}{7}(2) + (-\frac{4}{7})(-6) \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{8}{7} + \frac{18}{7} \\ \frac{6}{7} + \frac{24}{7} \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{26}{7} \\ \frac{30}{7} \end{bmatrix} \] By equality of matrices, \( x = \frac{26}{7}, y = \frac{30}{7} \) is the required solution. In simple words: The inversion method solves a system of linear equations by converting them into matrix form \( AX = B \), finding the inverse of matrix A (\( A^{-1} \)), and then calculating \( X = A^{-1}B \) to find the values of the variables.

🎯 Exam Tip: When using the matrix inversion method, pay close attention to the row operations to correctly transform matrix A into the identity matrix and obtain the accurate \( A^{-1} \).

Question 18. Solve the following equations by method of inversion: (ii) \( x + y - z = 2 \), \( x - 2y + z = 3 \) and \( 2x - y - 3z = -1 \) Solution: The given equations can be written in matrix form as: \[ \begin{bmatrix} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 2 & -1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \] This is of the form \( AX = B \), where \[ A = \begin{bmatrix} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 2 & -1 & -3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \] Let us find \( A^{-1} \). \( |A| = \begin{vmatrix} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 2 & -1 & -3 \end{vmatrix} \) \( = 1(6+1) - 1(-3-2) - 1(-1+4) \) \( = 1(7) - 1(-5) - 1(3) \) \( = 7+5-3 = 9 \neq 0 \)
Therefore, \( A^{-1} \) exists. We write \( AA^{-1}=I \)
Therefore, \[ \begin{bmatrix} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 2 & -1 & -3 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] By \( R_2 - R_1 \) and \( R_3 - 2R_1 \), we get \[ \begin{bmatrix} 1 & 1 & -1 \\ 0 & -3 & 2 \\ 0 & -3 & -1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \] By \( (-\frac{1}{3})R_2 \), we get \[ \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -\frac{2}{3} \\ 0 & -3 & -1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{3} & -\frac{1}{3} & 0 \\ -2 & 0 & 1 \end{bmatrix} \] By \( R_1 - R_2 \) and \( R_3 + 3R_2 \), we get \[ \begin{bmatrix} 1 & 0 & -\frac{1}{3} \\ 0 & 1 & -\frac{2}{3} \\ 0 & 0 & -3 \end{bmatrix} A^{-1} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & -\frac{1}{3} & 0 \\ -1 & -1 & 1 \end{bmatrix} \] By \( (-\frac{1}{3})R_3 \), we get \[ \begin{bmatrix} 1 & 0 & -\frac{1}{3} \\ 0 & 1 & -\frac{2}{3} \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \] By \( R_1 + \frac{1}{3}R_3 \) and \( R_2 + \frac{2}{3}R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} \frac{2}{3}+\frac{1}{9} & \frac{1}{3}+\frac{1}{9} & 0-\frac{1}{9} \\ \frac{1}{3}+\frac{2}{9} & -\frac{1}{3}+\frac{2}{9} & 0-\frac{2}{9} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \end{bmatrix} \] \[ A^{-1} = \begin{bmatrix} \frac{7}{9} & \frac{4}{9} & -\frac{1}{9} \\ \frac{5}{9} & -\frac{1}{9} & -\frac{2}{9} \\ \frac{3}{9} & \frac{3}{9} & -\frac{3}{9} \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 7 & 4 & -1 \\ 5 & -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} \] Now, premultiply \( AX = B \) by \( A^{-1} \), we get \( A^{-1}(AX) = A^{-1}B \)
\( \implies (A^{-1}A)X = A^{-1}B \)
\( \implies IX = A^{-1}B \)
Therefore, \( X = A^{-1}B \)
Therefore, \[ X = \frac{1}{9} \begin{bmatrix} 7 & 4 & -1 \\ 5 & -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 7(2)+4(3)+(-1)(-1) \\ 5(2)+(-1)(3)+(-2)(-1) \\ 3(2)+3(3)+(-3)(-1) \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 14+12+1 \\ 10-3+2 \\ 6+9+3 \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 27 \\ 9 \\ 18 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \] By equality of matrices, \( x = 3, y = 1, z = 2 \) is the required solution. In simple words: This method involves representing the system of three linear equations as a matrix equation, finding the inverse of the coefficient matrix using row operations, and then multiplying the inverse by the constant matrix to find the values of \(x\), \(y\), and \(z\).

🎯 Exam Tip: For 3x3 matrices, inversion by elementary operations can be lengthy. Keep track of fractions and simplify them at each step to minimize computational errors and ensure accuracy.

Question 18. Solve the following equations by method of inversion: (iii) \( x - y + z = 4 \), \( 2x + y - 3z = 0 \) and \( x + y + z = 2 \) Solution: Matrix form of the given system of equations is \[ \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \] This is of the form \( AX = B \), where \[ A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \] To determine \( X \), we have to find \( A^{-1} \). \( |A| = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} \) \( = 1(1 + 3) + 1(2 + 3) + 1(2-1) \) \( = 1(4) + 1(5) + 1(1) \) \( = 4+5+1 \) \( = 10 \neq 0 \)
Therefore, \( A^{-1} \) exists. Consider \( AA^{-1} = I \) \[ \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Applying \( R_2 \to R_2-2R_1 \) and \( R_3 \to R_3 - R_1 \), we get \[ \begin{bmatrix} 1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \] Applying \( R_2 \to (\frac{1}{3})R_2 \), we get \[ \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -\frac{5}{3} \\ 0 & 2 & 0 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -\frac{2}{3} & \frac{1}{3} & 0 \\ -1 & 0 & 1 \end{bmatrix} \] Applying \( R_1 \to R_1 + R_2 \) and \( R_3 \to R_3 - 2R_2 \), we get \[ \begin{bmatrix} 1 & 0 & -\frac{2}{3} \\ 0 & 1 & -\frac{5}{3} \\ 0 & 0 & \frac{10}{3} \end{bmatrix} A^{-1} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & 0 \\ -\frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & -\frac{2}{3} & 1 \end{bmatrix} \] Applying \( R_3 \to (\frac{3}{10})R_3 \), we get \[ \begin{bmatrix} 1 & 0 & -\frac{2}{3} \\ 0 & 1 & -\frac{5}{3} \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & 0 \\ -\frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{10} & -\frac{2}{10} & \frac{3}{10} \end{bmatrix} \] Applying \( R_1 \to R_1 + (\frac{2}{3})R_3 \) and \( R_2 \to R_2 + (\frac{5}{3})R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} \frac{1}{3}+\frac{2}{30} & \frac{1}{3}-\frac{4}{30} & 0+\frac{6}{30} \\ -\frac{2}{3}+\frac{5}{30} & \frac{1}{3}-\frac{10}{30} & 0+\frac{15}{30} \\ \frac{1}{10} & -\frac{2}{10} & \frac{3}{10} \end{bmatrix} \] \[ A^{-1} = \begin{bmatrix} \frac{12}{30} & \frac{6}{30} & \frac{6}{30} \\ -\frac{15}{30} & 0 & \frac{15}{30} \\ \frac{3}{30} & -\frac{6}{30} & \frac{9}{30} \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \] Pre-multiplying \( AX = B \) by \( A^{-1} \), we get \( A^{-1}(AX) = A^{-1}B \)
\( \implies (A^{-1}A) X = A^{-1}B \)
\( \implies IX = A^{-1}B \)
Therefore, \( X = A^{-1}B \)
Therefore, \[ X = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 4(4)+2(0)+2(2) \\ -5(4)+0(0)+5(2) \\ 1(4)-2(0)+3(2) \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4-0+6 \end{bmatrix} \] \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \]
Therefore, By equality of matrices, we get \( x = 2, y = -1 \) and \( z = 1 \). In simple words: This solution uses the matrix inversion method by first converting the system of linear equations into matrix form \( AX=B \), then performing elementary row operations to find \( A^{-1} \), and finally multiplying \( A^{-1} \) by \( B \) to determine the values of \( x, y, \) and \( z \).

🎯 Exam Tip: Accuracy in arithmetic with fractions during row transformations is key. Organize your steps clearly to minimize errors and allow for easy verification if needed.

Question 19. Solve the following equations by method of reduction: (i) \( 2x + y = 5 \), \( 3x - 5y = -3 \) Solution: The given equation can be written in matrix form as: \[ \begin{bmatrix} 2 & 1 \\ 3 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ -3 \end{bmatrix} \] By \( R_2 - 5R_1 \), we get \[ \begin{bmatrix} 2 & 1 \\ -7 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ -28 \end{bmatrix} \]
Therefore, \[ \begin{bmatrix} 2x + y \\ -7x + 0 \end{bmatrix} = \begin{bmatrix} 5 \\ -28 \end{bmatrix} \] By equality of matrices, \( 2x + y = 5 \) .......(1) \( -7x = -28 \) ......(2) From (2), \( x = 4 \) Substituting \( x = 4 \) in (1), we get \( 2(4) + y = 5 \) \( 8 + y = 5 \)
Therefore, \( y = -3 \) Hence, \( x = 4 \) and \( y = -3 \) is the required solution. In simple words: The reduction method solves a system of linear equations by converting them into matrix form and then using elementary row operations to transform the coefficient matrix into an upper triangular form, making it easy to solve for variables through back-substitution.

🎯 Exam Tip: The goal in the reduction method is to make the lower triangle of the coefficient matrix zeros. Choose row operations strategically to achieve this efficiently without introducing unnecessary fractions too early.

Question 19. Solve the following equations by method of reduction: (ii) \( x + 2y + z = 3 \), \( 3x - y + 2z = 1 \) and \( 2x - 3y + 3z = 2 \) Solution: The given equations can be written in matrix form as: \[ \begin{bmatrix} 1 & 2 & 1 \\ 3 & -1 & 2 \\ 2 & -3 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \] By \( R_2-3R_1 \) and \( R_3 -2R_1 \), we get \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & -7 & -1 \\ 0 & -7 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ -8 \\ -4 \end{bmatrix} \] By \( R_3-R_2 \), we get \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & -7 & -1 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ -8 \\ 4 \end{bmatrix} \]
Therefore, \[ \begin{bmatrix} x+2y+z \\ 0-7y-z \\ 0+0+2z \end{bmatrix} = \begin{bmatrix} 3 \\ -8 \\ 4 \end{bmatrix} \] By equality of matrices, \( x + 2y + z = 3 \) .......(1) \( -7y - z = -8 \) ......(2) \( 2z = 4 \) .......(3) From (3), \( z = 2 \) Substituting \( z = 2 \) in (2), we get \( -7y - 2 = -8 \)
Therefore, \( -7y = -6 \)
Therefore, \( y = \frac{6}{7} \) Substituting \( y = \frac{6}{7}, z = 2 \) in (1), we get \( x + 2(\frac{6}{7}) + 2 = 3 \) \( x = 3 - 2 - \frac{12}{7} = 1 - \frac{12}{7} = -\frac{5}{7} \) Hence, \( x = -\frac{5}{7}, y = \frac{6}{7} \) and \( z = 2 \) is the required solution. In simple words: This problem is solved using matrix reduction, where the system of equations is converted into a matrix, then row operations are applied to simplify the matrix into an upper triangular form, allowing for straightforward back-substitution to find the values of \(x\), \(y\), and \(z\).

🎯 Exam Tip: When performing row operations, ensure you apply the operation to all elements in the row, including the elements in the constant matrix B, to maintain consistency and correctness of the system.

Question 19. Solve the following equations by method of reduction: (iii) \( x - 3y + z = 2 \), \( 3x + y + z = 1 \) and \( 5x + y + 3z = 3 \). Solution: Matrix form of the given system of equations is \[ \begin{bmatrix} 1 & -3 & 1 \\ 3 & 1 & 1 \\ 5 & 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \] This is of the form \( AX = B \), where \[ A = \begin{bmatrix} 1 & -3 & 1 \\ 3 & 1 & 1 \\ 5 & 1 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \] Applying \( R_2 \to R_2-3R_1 \) and \( R_3 \to R_3 - 5R_1 \), we get \[ \begin{bmatrix} 1 & -3 & 1 \\ 0 & 10 & -2 \\ 0 & 16 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ -5 \\ -7 \end{bmatrix} \] Applying \( R_3 \to R_3-(\frac{8}{5})R_2 \), we get \[ \begin{bmatrix} 1 & -3 & 1 \\ 0 & 10 & -2 \\ 0 & 0 & \frac{6}{5} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ -5 \\ 1 \end{bmatrix} \] Hence, the original matrix A is reduced to an upper triangular matrix.
Therefore, \[ \begin{bmatrix} x - 3y + z \\ 0 + 10y - 2z \\ 0 + 0 + \frac{6}{5}z \end{bmatrix} = \begin{bmatrix} 2 \\ -5 \\ 1 \end{bmatrix} \]
Therefore, By equality of matrices, we get \( x - 3y + z = 2 \) ...(i) \( 10y - 2z = -5 \) ...(ii) \( \frac{6}{5}z = 1 \)
Therefore, \( z = \frac{5}{6} \) Substituting \( z = \frac{5}{6} \) in equation (ii), we get \( 10y - 2(\frac{5}{6}) = -5 \) \( 10y - \frac{10}{6} = -5 \)
Therefore, \( 10y = -5 + \frac{10}{6} = \frac{-30+10}{6} = -\frac{20}{6} \)
Therefore, \( 10y = -\frac{10}{3} \)
Therefore, \( y = -\frac{1}{3} \) Substituting \( y = -\frac{1}{3} \) and \( z = \frac{5}{6} \) in equation (i), we get \( x - 3(-\frac{1}{3}) + \frac{5}{6} = 2 \) \( x + 1 + \frac{5}{6} = 2 \)
Therefore, \( x = 2 - 1 - \frac{5}{6} = 1 - \frac{5}{6} = \frac{1}{6} \)
Therefore, \( x = \frac{1}{6}, y = -\frac{1}{3} \) and \( z = \frac{5}{6} \) is the required solution. In simple words: This solution uses the matrix reduction method by converting the system of equations into a matrix, then applying row operations to achieve an upper triangular form. This simplified matrix allows for sequential solving of variables through back-substitution.

🎯 Exam Tip: Always verify your solution by substituting the calculated values of \(x\), \(y\), and \(z\) back into the original equations to ensure they are all satisfied. This helps catch any calculation errors.

Question 20. The sum of three numbers is 6. If we multiply the third number by 3 and add it to the second number, we get 11. By adding first and third numbers we get a number that is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices. Solution: Let the three numbers be \( x, y, \) and \( z \). According to the given condition, \( x + y + z = 6 \) \( 3z + y = 11 \), i.e. \( y + 3z = 11 \) and \( x + z = 2y \), i.e. \( x - 2y + z = 0 \) Hence, the system of linear equations is \( x + y + z = 6 \) \( y + 3z = 11 \) \( x - 2y + z = 0 \) These equations can be written in matrix form as: \[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} \] By \( R_3 \to R_3 - R_1 \), we get \[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ -6 \end{bmatrix} \]
Therefore, \[ \begin{bmatrix} x+y+z \\ 0+y+3z \\ 0-3y+0 \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ -6 \end{bmatrix} \] By equality of matrices, \( x + y + z = 6 \) .......(1) \( y + 3z = 11 \) ......(2) \( -3y = -6 \) ......(3) From (3), \( y = 2 \) Substituting \( y = 2 \) in (2), we get \( 2 + 3z = 11 \)
Therefore, \( 3z = 9 \)
Therefore, \( z = 3 \) Substituting \( y = 2, z = 3 \) in (1), we get \( x + 2 + 3 = 6 \)
Therefore, \( x + 5 = 6 \)
Therefore, \( x = 1 \)
Therefore, \( x = 1, y = 2, z = 3 \) Hence, the required numbers are 1, 2 and 3. In simple words: This problem translates a word problem into a system of linear equations, which is then solved using the matrix reduction method. The matrix is transformed to an upper triangular form via row operations, allowing the variables to be found by back-substitution.

🎯 Exam Tip: Accurately translating the word problem into correct linear equations is the first and most critical step. Any error here will propagate throughout the matrix solution, so review this initial setup carefully.