Samacheer Kalvi Class 11 Maths Solutions Chapter 7 Matrices and Determinants Exercise 7.1

Question 1. Construct an m × n matrix A = [a_ij], where a_ij is given by
(i) a_ij = \( \frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2} \) with m = 2, n = 3
(ii) a_ij = \( \frac{|3 \mathbf{i}-4 \mathbf{j}|}{4} \) with m = 3, n = 4
Answer:
(i) To construct a 2 x 3 matrix, we need elements for A = \( \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \). We use the formula \( a_{ij} = \frac{(i-2j)^{2}}{2} \) to find each element.
\( a_{11} = \frac{(1-2 \times 1)^{2}}{2} = \frac{(-1)^{2}}{2} = \frac{1}{2} \)
\( a_{12} = \frac{(1-2 \times 2)^{2}}{2} = \frac{(-3)^{2}}{2} = \frac{9}{2} \)
\( a_{13} = \frac{(1-2 \times 3)^{2}}{2} = \frac{(-5)^{2}}{2} = \frac{25}{2} \)
\( a_{21} = \frac{(2-2 \times 1)^{2}}{2} = \frac{(0)^{2}}{2} = 0 \)
\( a_{22} = \frac{(2-2 \times 2)^{2}}{2} = \frac{(-2)^{2}}{2} = 2 \)
\( a_{23} = \frac{(2-2 \times 3)^{2}}{2} = \frac{(-4)^{2}}{2} = 8 \)
The resulting 2 x 3 matrix A is: \[ A = \begin{bmatrix} \frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\ 0 & 2 & 8 \end{bmatrix} \]
(ii) To construct a 3 x 4 matrix, we need elements for A = \( \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix} \). We use the formula \( a_{ij} = \frac{|3i-4j|}{4} \) to find each element. The absolute value ensures all elements are non-negative before division.
\( a_{11} = \frac{|3 \times 1-4 \times 1|}{4} = \frac{|-1|}{4} = \frac{1}{4} \)
\( a_{12} = \frac{|3 \times 1-4 \times 2|}{4} = \frac{|-5|}{4} = \frac{5}{4} \)
\( a_{13} = \frac{|3 \times 1-4 \times 3|}{4} = \frac{|-9|}{4} = \frac{9}{4} \)
\( a_{14} = \frac{|3 \times 1-4 \times 4|}{4} = \frac{|-13|}{4} = \frac{13}{4} \)
\( a_{21} = \frac{|3 \times 2-4 \times 1|}{4} = \frac{|2|}{4} = \frac{1}{2} \)
\( a_{22} = \frac{|3 \times 2-4 \times 2|}{4} = \frac{|-2|}{4} = \frac{1}{2} \)
\( a_{23} = \frac{|3 \times 2-4 \times 3|}{4} = \frac{|-6|}{4} = \frac{3}{2} \)
\( a_{24} = \frac{|3 \times 2-4 \times 4|}{4} = \frac{|-10|}{4} = \frac{5}{2} \)
\( a_{31} = \frac{|3 \times 3-4 \times 1|}{4} = \frac{|5|}{4} = \frac{5}{4} \)
\( a_{32} = \frac{|3 \times 3-4 \times 2|}{4} = \frac{|1|}{4} = \frac{1}{4} \)
\( a_{33} = \frac{|3 \times 3-4 \times 3|}{4} = \frac{|-3|}{4} = \frac{3}{4} \)
\( a_{34} = \frac{|3 \times 3-4 \times 4|}{4} = \frac{|-7|}{4} = \frac{7}{4} \)
The resulting 3 x 4 matrix A is: \[ A = \begin{bmatrix} \frac{1}{4} & \frac{5}{4} & \frac{9}{4} & \frac{13}{4} \\ \frac{1}{2} & \frac{1}{2} & \frac{3}{2} & \frac{5}{2} \\ \frac{5}{4} & \frac{1}{4} & \frac{3}{4} & \frac{7}{4} \end{bmatrix} \]
In simple words: To build a matrix, you fill in each spot using a given rule, which depends on the row and column number. Then you write out the matrix with all the calculated numbers in their places. Matrices are fundamental for solving systems of linear equations.

🎯 Exam Tip: Always double-check your calculations for each element, especially when dealing with squares or absolute values, as small errors can lead to an incorrect final matrix.

Question 2. Find the value of p, q, r and s if
\[ \begin{bmatrix} p^{2}-1 & 0 & -31-q^{3} \\ 7 & r+1 & 9 \\ -2 & 8 & s-1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -4 \\ 7 & \frac{3}{2} & 9 \\ -2 & 8 & -\pi \end{bmatrix} \]
Answer: To find the values of p, q, r, and s, we set the corresponding elements of the two matrices equal to each other. This is because two matrices are equal if and only if all their corresponding elements are equal.
From the first row, first column: \( p^{2}-1 = 1 \)
\( p^{2} = 1 + 1 \)
\( p^{2} = 2 \)
\( p = \pm \sqrt{2} \)

From the first row, third column: \( -31-q^{3} = -4 \)
\( -q^{3} = -4 + 31 \)
\( -q^{3} = 27 \)
\( q^{3} = -27 \)
\( q^{3} = (-3)^{3} \)
\( \implies q = -3 \)

From the second row, second column: \( r+1 = \frac{3}{2} \)
\( r = \frac{3}{2} - 1 \)
\( r = \frac{3-2}{2} \)
\( r = \frac{1}{2} \)

From the third row, third column: \( s-1 = -\pi \)
\( s = -\pi + 1 \)
So, the values are \( p = \pm \sqrt{2} \), \( q = -3 \), \( r = \frac{1}{2} \) and \( s = 1-\pi \).
In simple words: When two matrices are exactly the same, it means every number in the same spot in both matrices must be equal. We use this idea to solve for the unknown letters.

🎯 Exam Tip: Remember to consider both positive and negative square roots for variables like \( p^2 \) to ensure all possible solutions are included.

Question 3. Determine the value of x + y if
\[ \begin{bmatrix} 2x + y & 4x \\ 5x - 7 & 4x \end{bmatrix} = \begin{bmatrix} 7 & 7y - 13 \\ y & x + 6 \end{bmatrix} \]
Answer: Since the two matrices are equal, their corresponding elements must be equal. We can set up equations by matching the elements in the same positions.
Equating the element in the third row, third column (This seems like a typo, the matrices are 2x2. It should be second row, second column):
\( 4x = x + 6 \)
\( 4x - x = 6 \)
\( 3x = 6 \)
\( x = \frac{6}{3} \)
\( x = 2 \)

Now, we use this value of \(x\) to find \(y\). Let's use the first row, first column element equation:
\( 2x + y = 7 \)
Substitute \( x = 2 \):
\( 2(2) + y = 7 \)
\( 4 + y = 7 \)
\( y = 7 - 4 \)
\( y = 3 \)

We can also verify with other elements. For example, second row, first column:
\( 5x - 7 = y \)
\( 5(2) - 7 = 3 \)
\( 10 - 7 = 3 \)
\( 3 = 3 \) (This matches!)

Now, calculate \(x + y\):
\( x + y = 2 + 3 \)
\( x + y = 5 \)
In simple words: We find the values of x and y by matching the numbers in the same spots in both matrices. We solve these mini-equations to find x, then use x to find y, and finally add them up.

🎯 Exam Tip: When you have multiple equations, always start by solving the simplest one that has only one variable, like \(4x = x+6\), before moving to more complex equations.

Question 4. Determine the matrices A and B if they satisfy \( 2A - B + \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} = 0 \) and \( A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \).
Answer: We are given two matrix equations. We can solve them like a system of linear equations, but with matrices.
The first equation is: \( 2A - B + \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
This can be rewritten as: \( 2A - B = - \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \)
So, \( 2A - B = \begin{bmatrix} -6 & 6 & 0 \\ 4 & -2 & -1 \end{bmatrix} \) (Equation 1)

The second equation is: \( A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \) (Equation 2)

To eliminate A, multiply Equation 1 by 2:
\( 2(2A - B) = 2 \begin{bmatrix} -6 & 6 & 0 \\ 4 & -2 & -1 \end{bmatrix} \)
\( 4A - 2B = \begin{bmatrix} -12 & 12 & 0 \\ 8 & -4 & -2 \end{bmatrix} \) (Equation 3)

Now, subtract Equation 2 from Equation 3:
\( (4A - 2B) - (A - 2B) = \begin{bmatrix} -12 & 12 & 0 \\ 8 & -4 & -2 \end{bmatrix} - \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)
\( 3A = \begin{bmatrix} -12-3 & 12-2 & 0-8 \\ 8-(-2) & -4-1 & -2-(-7) \end{bmatrix} \)
\( 3A = \begin{bmatrix} -15 & 10 & -8 \\ 10 & -5 & 5 \end{bmatrix} \)
\( A = \frac{1}{3} \begin{bmatrix} -15 & 10 & -8 \\ 10 & -5 & 5 \end{bmatrix} \)
\[ A = \begin{bmatrix} -5 & \frac{10}{3} & -\frac{8}{3} \\ \frac{10}{3} & -\frac{5}{3} & \frac{5}{3} \end{bmatrix} \]

Next, substitute A back into Equation 2 to find B:
\( 2B = A - \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)
\[ 2B = \begin{bmatrix} -5 & \frac{10}{3} & -\frac{8}{3} \\ \frac{10}{3} & -\frac{5}{3} & \frac{5}{3} \end{bmatrix} - \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \]
\[ 2B = \begin{bmatrix} -5-3 & \frac{10}{3}-2 & -\frac{8}{3}-8 \\ \frac{10}{3}-(-2) & -\frac{5}{3}-1 & \frac{5}{3}-(-7) \end{bmatrix} \]
\[ 2B = \begin{bmatrix} -8 & \frac{10-6}{3} & \frac{-8-24}{3} \\ \frac{10+6}{3} & \frac{-5-3}{3} & \frac{5+21}{3} \end{bmatrix} \]
\[ 2B = \begin{bmatrix} -8 & \frac{4}{3} & -\frac{32}{3} \\ \frac{16}{3} & -\frac{8}{3} & \frac{26}{3} \end{bmatrix} \]
\[ B = \frac{1}{2} \begin{bmatrix} -8 & \frac{4}{3} & -\frac{32}{3} \\ \frac{16}{3} & -\frac{8}{3} & \frac{26}{3} \end{bmatrix} \]
\[ B = \begin{bmatrix} -4 & \frac{2}{3} & -\frac{16}{3} \\ \frac{8}{3} & -\frac{4}{3} & \frac{13}{3} \end{bmatrix} \]
In simple words: We treat the matrix equations like regular equations for numbers. We multiply and subtract the whole matrices to get rid of one unknown matrix, then solve for the other. Matrix operations are done element by element.

🎯 Exam Tip: When manipulating matrix equations, remember to distribute scalar multiplication to all elements within the matrix and perform addition/subtraction element-wise. Pay close attention to signs, especially when subtracting matrices.

Question 5. If \( A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \), then compute \( A^4 \).
Answer: We need to find \( A^4 \), which means multiplying matrix A by itself four times. We can do this in steps: first find \( A^2 \), then multiply \( A^2 \) by itself to get \( A^4 \).
Given matrix \( A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \).

First, calculate \( A^2 \):
\( A^2 = A \cdot A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \)
To multiply matrices, we multiply rows by columns:
\( A^2 = \begin{bmatrix} (1 \times 1) + (a \times 0) & (1 \times a) + (a \times 1) \\ (0 \times 1) + (1 \times 0) & (0 \times a) + (1 \times 1) \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 1+0 & a+a \\ 0+0 & 0+1 \end{bmatrix} \)
\[ A^2 = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \]

Now, calculate \( A^4 \) using \( A^4 = A^2 \cdot A^2 \):
\( A^4 = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \)
Again, multiply rows by columns:
\( A^4 = \begin{bmatrix} (1 \times 1) + (2a \times 0) & (1 \times 2a) + (2a \times 1) \\ (0 \times 1) + (1 \times 0) & (0 \times 2a) + (1 \times 1) \end{bmatrix} \)
\( A^4 = \begin{bmatrix} 1+0 & 2a+2a \\ 0+0 & 0+1 \end{bmatrix} \)
\[ A^4 = \begin{bmatrix} 1 & 4a \\ 0 & 1 \end{bmatrix} \]
Notice a pattern: for this type of matrix, \(A^n\) would be \( \begin{bmatrix} 1 & na \\ 0 & 1 \end{bmatrix} \).
In simple words: To find a matrix raised to a power, we multiply the matrix by itself that many times. First, we found A squared, then we multiplied A squared by itself to get A to the power of four.

🎯 Exam Tip: For problems involving higher powers of matrices, calculating \(A^2\) and \(A^3\) often reveals a pattern, which can simplify finding \(A^n\).

Question 6. Consider the matrix \( A_\alpha = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \).
(i) Show that \( A_\alpha A_\beta = A_{(\alpha+\beta)} \)
(ii) Find all possible real values \( \alpha \) satisfying the condition \( A_\alpha + A_\alpha^T = I \).
Answer:
(i) First, let's write down the matrix \( A_\beta \):
\( A_\beta = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix} \)

Now, we multiply \( A_\alpha \) by \( A_\beta \):
\( A_\alpha A_\beta = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix} \)
\[ A_\alpha A_\beta = \begin{bmatrix} (\cos \alpha)(\cos \beta) - (\sin \alpha)(\sin \beta) & (\cos \alpha)(-\sin \beta) - (\sin \alpha)(\cos \beta) \\ (\sin \alpha)(\cos \beta) + (\cos \alpha)(\sin \beta) & (\sin \alpha)(-\sin \beta) + (\cos \alpha)(\cos \beta) \end{bmatrix} \]
Using the trigonometric identities for sum of angles:
\( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y \)
\( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \)
The matrix simplifies to:
\[ A_\alpha A_\beta = \begin{bmatrix} \cos(\alpha+\beta) & -(\sin \beta \cos \alpha + \cos \beta \sin \alpha) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix} \]
\[ A_\alpha A_\beta = \begin{bmatrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix} \]
This is exactly the definition of \( A_{(\alpha+\beta)} \). Therefore, \( A_\alpha A_\beta = A_{(\alpha+\beta)} \) is shown.

(ii) We need to find the real values of \( \alpha \) for which \( A_\alpha + A_\alpha^T = I \).
First, let's find the transpose of \( A_\alpha \), which is \( A_\alpha^T \):
\[ A_\alpha^T = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \]

Now, we add \( A_\alpha \) and \( A_\alpha^T \):
\( A_\alpha + A_\alpha^T = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} + \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)
\[ A_\alpha + A_\alpha^T = \begin{bmatrix} \cos \alpha + \cos \alpha & -\sin \alpha + \sin \alpha \\ \sin \alpha - \sin \alpha & \cos \alpha + \cos \alpha \end{bmatrix} \]
\[ A_\alpha + A_\alpha^T = \begin{bmatrix} 2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha \end{bmatrix} \]

The identity matrix \( I \) for a 2x2 matrix is \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
So, we set the sum equal to the identity matrix:
\[ \begin{bmatrix} 2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
By equating the corresponding elements, we get:
\( 2 \cos \alpha = 1 \)
\( \cos \alpha = \frac{1}{2} \)
The general solution for \( \cos \alpha = \frac{1}{2} \) is \( \alpha = 2n\pi \pm \frac{\pi}{3} \), where \( n \in Z \). Rotational matrices are very important in computer graphics and physics.
In simple words: For part (i), we multiply two matrices that use angles, and the answer shows that the angles add up inside the new matrix, just like expected. For part (ii), we add a matrix to its "flipped" version (transpose) and make it equal to a special "identity" matrix. This helps us find what the angle must be.

🎯 Exam Tip: When proving matrix identities involving trigonometric functions, remember to apply the relevant sum and difference formulas for sine and cosine. For equations, ensure you find the general solution for trigonometric functions.

Question 7. If \( A = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} \) and such that \( (A - 2I)(A - 3I) = 0 \), find the value of \( x \).
Answer: We are given a matrix A and a condition involving the identity matrix I. We need to find the value of \( x \) that makes the product of two derived matrices equal to the zero matrix.
Given matrix \( A = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} \). The identity matrix \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).

First, calculate \( (A - 2I) \):
\( A - 2I = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( A - 2I = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \)
\[ A - 2I = \begin{bmatrix} 4-2 & 2-0 \\ -1-0 & x-2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -1 & x-2 \end{bmatrix} \]

Next, calculate \( (A - 3I) \):
\( A - 3I = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( A - 3I = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\[ A - 3I = \begin{bmatrix} 4-3 & 2-0 \\ -1-0 & x-3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & x-3 \end{bmatrix} \]

Now, multiply \( (A - 2I) \) by \( (A - 3I) \):
\( (A - 2I)(A - 3I) = \begin{bmatrix} 2 & 2 \\ -1 & x-2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & x-3 \end{bmatrix} \)
\[ (A - 2I)(A - 3I) = \begin{bmatrix} (2)(1)+(2)(-1) & (2)(2)+(2)(x-3) \\ (-1)(1)+(x-2)(-1) & (-1)(2)+(x-2)(x-3) \end{bmatrix} \]
\[ (A - 2I)(A - 3I) = \begin{bmatrix} 2-2 & 4+2x-6 \\ -1-(x-2) & -2+(x^2-3x-2x+6) \end{bmatrix} \]
\[ (A - 2I)(A - 3I) = \begin{bmatrix} 0 & 2x-2 \\ -1-x+2 & -2+x^2-5x+6 \end{bmatrix} \]
\[ (A - 2I)(A - 3I) = \begin{bmatrix} 0 & 2x-2 \\ 1-x & x^2-5x+4 \end{bmatrix} \]

We are given that \( (A - 2I)(A - 3I) = 0 \), where \( 0 \) is the zero matrix \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
So, we equate the elements of the resulting matrix to zero:
\( 2x - 2 = 0 \implies 2x = 2 \implies x = 1 \)
\( 1 - x = 0 \implies 1 = x \implies x = 1 \)
\( x^2 - 5x + 4 = 0 \)
Let's check if \( x=1 \) satisfies this quadratic equation: \( (1)^2 - 5(1) + 4 = 1 - 5 + 4 = 0 \). It does.
The required value of \( x \) is \( 1 \). This type of equation is often seen in finding eigenvalues of a matrix.
In simple words: First, we subtract identity matrices from A, then we multiply the results. Since the final answer should be a matrix of all zeros, we set the parts with x equal to zero to find x.

🎯 Exam Tip: Remember that matrix multiplication is not commutative, so the order of \((A-2I)\) and \((A-3I)\) matters. Also, be careful with signs when subtracting matrices and expanding quadratic expressions.

Question 7. If \(A = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix}\) and such that \((A – 2I) (A – 3I) = 0\), find the value of x.
Answer: First, we will find \(A-2I\). This means we subtract two times the identity matrix from matrix A.
\(A - 2I = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\( = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
\( = \begin{bmatrix} 4-2 & 2-0 \\ -1-0 & x-2 \end{bmatrix}\)
\( = \begin{bmatrix} 2 & 2 \\ -1 & x-2 \end{bmatrix}\)
Next, we will find \(A-3I\). This means we subtract three times the identity matrix from matrix A.
\(A - 3I = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\( = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
\( = \begin{bmatrix} 4-3 & 2-0 \\ -1-0 & x-3 \end{bmatrix}\)
\( = \begin{bmatrix} 1 & 2 \\ -1 & x-3 \end{bmatrix}\)
Now, we multiply the two resulting matrices, \((A-2I)(A-3I)\).
\((A-2I)(A-3I) = \begin{bmatrix} 2 & 2 \\ -1 & x-2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & x-3 \end{bmatrix}\)
\( = \begin{bmatrix} 2(1)+2(-1) & 2(2)+2(x-3) \\ -1(1)+(x-2)(-1) & -1(2)+(x-2)(x-3) \end{bmatrix}\)
\( = \begin{bmatrix} 2-2 & 4+2x-6 \\ -1-x+2 & -2+x^2-3x-2x+6 \end{bmatrix}\)
\( = \begin{bmatrix} 0 & 2x-2 \\ 1-x & x^2-5x+4 \end{bmatrix}\)
We are given that \((A-2I)(A-3I) = 0\), where 0 represents the zero matrix \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
So, we can equate the entries of the resulting matrix to the zero matrix.
From the top right entry: \(2x-2 = 0\)
\( \implies 2x = 2 \)
\( \implies x = 1 \)
From the bottom right entry: \(x^2-5x+4 = 0\)
To check if \(x=1\) satisfies this, substitute \(x=1\):
\( (1)^2-5(1)+4 = 1-5+4 = 0 \)
Since both equations give \(x=1\), this is the correct value. This method helps confirm the answer by using different parts of the matrix. Therefore, the required value of x is 1.
In simple words: First, calculate the two matrices \(A-2I\) and \(A-3I\). Then, multiply these two new matrices together. Since the result should be a matrix of all zeros, set each part of the multiplied matrix to zero. Solving these simple equations will give you the value of x.

🎯 Exam Tip: Always verify your value of x by substituting it back into all equations derived from the matrix equality to ensure consistency and avoid errors.

Question 8. If \(A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}\), show that \(A^2\) is a unit matrix.
Answer: To find \(A^2\), we need to multiply matrix A by itself, so \(A^2 = A \cdot A\). A unit matrix, also known as an identity matrix, has ones on its main diagonal and zeros everywhere else.
\(A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix}\)
\( = \begin{bmatrix} (1)(1)+(0)(0)+(0)(a) & (1)(0)+(0)(1)+(0)(b) & (1)(0)+(0)(0)+(0)(-1) \\ (0)(1)+(1)(0)+(0)(a) & (0)(0)+(1)(1)+(0)(b) & (0)(0)+(1)(0)+(0)(-1) \\ (a)(1)+(b)(0)+(-1)(a) & (a)(0)+(b)(1)+(-1)(b) & (a)(0)+(b)(0)+(-1)(-1) \end{bmatrix}\)
\( = \begin{bmatrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ a+0-a & 0+b-b & 0+0+1 \end{bmatrix}\)
\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
This resulting matrix is the identity matrix of order 3. An identity matrix acts like the number 1 in multiplication for matrices, leaving other matrices unchanged. Thus, \(A^2\) is a unit matrix.
In simple words: Multiply matrix A by itself. When you do the math, all the values on the main diagonal will be 1, and all other values will be 0. This is the definition of a unit (or identity) matrix.

🎯 Exam Tip: Remember that for an identity matrix I, \(I \cdot A = A \cdot I = A\). Identifying identity matrices quickly can simplify calculations.

Question 9. If \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\) and \(A^3 – 6A^2 + 7A + kI = 0\), find the value of k.
Answer: We need to find \(A^2\), \(A^3\), and then substitute them into the given equation to solve for k. An identity matrix I is used here, which has ones on the main diagonal and zeros elsewhere. First, calculate \(A^2\):
\(A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\)
\( = \begin{bmatrix} 1(1)+0(0)+2(2) & 1(0)+0(2)+2(0) & 1(2)+0(1)+2(3) \\ 0(1)+2(0)+1(2) & 0(0)+2(2)+1(0) & 0(2)+2(1)+1(3) \\ 2(1)+0(0)+3(2) & 2(0)+0(2)+3(0) & 2(2)+0(1)+3(3) \end{bmatrix}\)
\( = \begin{bmatrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{bmatrix}\)
\( = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}\)
Next, calculate \(A^3\):
\(A^3 = A^2 \cdot A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\)
\( = \begin{bmatrix} 5(1)+0(0)+8(2) & 5(0)+0(2)+8(0) & 5(2)+0(1)+8(3) \\ 2(1)+4(0)+5(2) & 2(0)+4(2)+5(0) & 2(2)+4(1)+5(3) \\ 8(1)+0(0)+13(2) & 8(0)+0(2)+13(0) & 8(2)+0(1)+13(3) \end{bmatrix}\)
\( = \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{bmatrix}\)
\( = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}\)
Now, substitute \(A^3\), \(A^2\), A, and I into the given equation \(A^3 – 6A^2 + 7A + kI = 0\):
\( \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6 \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + k \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix} + \begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
Combine the matrices:
\( \begin{bmatrix} 21-30+7+k & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+k & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+k \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} -2+k & 0 & 0 \\ 0 & -2+k & 0 \\ 0 & 0 & -2+k \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
Equating any corresponding entry to 0, for example, the top left entry:
\( -2+k = 0 \)
\( \implies k = 2 \)
The value of k is 2. This process is often used to demonstrate the Cayley-Hamilton theorem, where a matrix satisfies its own characteristic equation.
In simple words: First, calculate \(A^2\) (A times A) and \(A^3\) (A times \(A^2\)). Then, put all these matrices into the given equation, remembering that I is an identity matrix. Add and subtract all the parts. Finally, set any part of the resulting matrix to zero to find the value of k.

🎯 Exam Tip: Be careful with matrix multiplication, as it's the most common source of errors. Always double-check your calculations, especially when combining several terms.

Question 10. Give your own examples of matrices satisfying the following conditions in each case:
(i) A and B such that AB \(\neq\) BA
(ii) A and B such that AB = 0 = BA, A \(\neq\) 0 and B \(\neq\) 0.
(iii) A and B such that AB = 0 and BA \(\neq\) 0
Answer: We will provide examples for each condition. The concept of matrix multiplication not being commutative (AB \(\neq\) BA) is a fundamental difference from scalar multiplication.
(i) A and B such that AB \(\neq\) BA
Let \(A = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\)
First, calculate AB:
\(AB = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 1(3)+2(5) & 1(1)+2(2) \\ 4(3)+5(5) & 4(1)+5(2) \end{bmatrix} = \begin{bmatrix} 3+10 & 1+4 \\ 12+25 & 4+10 \end{bmatrix} = \begin{bmatrix} 13 & 5 \\ 37 & 14 \end{bmatrix}\)
Next, calculate BA:
\(BA = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 3(1)+1(4) & 3(2)+1(5) \\ 5(1)+2(4) & 5(2)+2(5) \end{bmatrix} = \begin{bmatrix} 3+4 & 6+5 \\ 5+8 & 10+10 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ 13 & 20 \end{bmatrix}\)
Since \(\begin{bmatrix} 13 & 5 \\ 37 & 14 \end{bmatrix} \neq \begin{bmatrix} 7 & 11 \\ 13 & 20 \end{bmatrix}\), we have AB \(\neq\) BA.
(ii) A and B such that AB = 0 = BA, A \(\neq\) 0 and B \(\neq\) 0.
Let \(A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\). Both A and B are non-zero matrices.
First, calculate AB:
\(AB = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+(-1)(1) & 1(1)+(-1)(1) \\ (-1)(1)+1(1) & (-1)(1)+1(1) \end{bmatrix} = \begin{bmatrix} 1-1 & 1-1 \\ -1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Next, calculate BA:
\(BA = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+1(-1) & 1(-1)+1(1) \\ 1(1)+1(-1) & 1(-1)+1(1) \end{bmatrix} = \begin{bmatrix} 1-1 & -1+1 \\ 1-1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Here, AB = 0 and BA = 0, and A \(\neq\) 0, B \(\neq\) 0.
(iii) A and B such that AB = 0 and BA \(\neq\) 0
Let \(A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\)
First, calculate AB:
\(AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0(1)+1(0) & 0(0)+1(0) \\ 0(1)+0(0) & 0(0)+0(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Next, calculate BA:
\(BA = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1(0)+0(0) & 1(1)+0(0) \\ 0(0)+0(0) & 0(1)+0(0) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)
Here, AB = 0 but BA \(\neq\) 0.
In simple words: This question asks for examples to show how matrices behave differently from regular numbers. For (i), you need two matrices that give different answers when multiplied in different orders. For (ii), find two matrices that are not zero themselves, but when you multiply them, the result is a zero matrix, no matter the order. For (iii), find two matrices where multiplying in one order gives a zero matrix, but multiplying in the opposite order does not.

🎯 Exam Tip: When providing examples, choose simple matrices with small integer entries to make calculations easier and clearer. Always verify your examples through multiplication.

Question 11. Show that \(f(x) f(y) = f(x + y)\), where \(f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
Answer: We need to calculate the product \(f(x)f(y)\) and show it is equal to \(f(x+y)\). This problem uses trigonometric identities for addition of angles.
Given \(f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
Therefore, \(f(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
And \(f(x+y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
Now, let's multiply \(f(x)\) and \(f(y)\):
\(f(x)f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
\( = \begin{bmatrix} (\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) & (\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) & (\cos x)(0) + (-\sin x)(0) + (0)(1) \\ (\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) & (\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) & (\sin x)(0) + (\cos x)(0) + (0)(1) \\ (0)(\cos y) + (0)(\sin y) + (1)(0) & (0)(-\sin y) + (0)(\cos y) + (1)(0) & (0)(0) + (0)(0) + (1)(1) \end{bmatrix}\)
\( = \begin{bmatrix} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
Using the trigonometric identities:
\(\cos(x+y) = \cos x \cos y - \sin x \sin y\)
\(\sin(x+y) = \sin x \cos y + \cos x \sin y\)
Substitute these identities into the product matrix:
\(f(x)f(y) = \begin{bmatrix} \cos(x+y) & -(\cos x \sin y + \sin x \cos y) & 0 \\ \sin(x+y) & \cos x \cos y - \sin x \sin y & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
\( = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
This is exactly \(f(x+y)\). Therefore, we have shown that \(f(x) f(y) = f(x + y)\). This kind of matrix is related to rotation matrices in geometry.
In simple words: Multiply the two matrices \(f(x)\) and \(f(y)\) together. When you finish the multiplication, use your knowledge of angle addition formulas from trigonometry. You will find that the new matrix looks exactly like \(f(x+y)\).

🎯 Exam Tip: Memorize the basic trigonometric addition formulas for sine and cosine, as they are essential for problems involving matrices with trigonometric entries.

Question 12. If A is a square matrix such that \(A^2 = A\), find the value of \(7A – (I + A)^3\).
Answer: We are given that \(A^2 = A\). A matrix that satisfies this condition is called an idempotent matrix. We need to simplify the expression \(7A – (I + A)^3\).
First, let's expand \((I + A)^3\) using the binomial expansion formula, similar to \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):
\((I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3\)
Since I is the identity matrix, \(I^3 = I\) and \(I^2 = I\). Also, \(IA = A\) and \(A^2 = A\).
Let's find \(A^3\):
\(A^3 = A^2 \cdot A = A \cdot A\) (since \(A^2 = A\))
\(A^3 = A^2 = A\)
Now substitute these into the expansion of \((I+A)^3\):
\((I + A)^3 = I + 3IA + 3IA^2 + A^3\)
\( = I + 3A + 3A + A \)
\( = I + 7A \)
Now, substitute this back into the original expression \(7A – (I + A)^3\):
\(7A – (I + A)^3 = 7A – (I + 7A)\)
\( = 7A – I – 7A \)
\( = -I \)
So, the value of \(7A – (I + A)^3\) is \(-I\), which is the negative of the identity matrix. This shows how matrix properties can simplify complex expressions.
In simple words: First, use the special rule \(A^2 = A\) to simplify \(A^3\). Then, expand \((I+A)^3\) just like you would with regular numbers, remembering that I is the identity matrix. Put all these simplified parts back into the main problem and do the subtractions. You will find that the answer is just \(-I\).

🎯 Exam Tip: For problems involving \(A^2 = A\), always remember that \(A^n = A\) for any positive integer n. This shortcut can save significant time in calculations.

Question 13. Verify the property \(A (B + C) = AB + AC\), when the matrices A, B, and C are given by \(A = \begin{bmatrix} 2 & 0 & -3 \\ 1 & 4 & 5 \end{bmatrix}\), \(B = \begin{bmatrix} 3 & 1 \\ -1 & 0 \\ 4 & 2 \end{bmatrix}\), and \(C = \begin{bmatrix} 4 & 7 \\ 2 & 1 \\ 1 & -1 \end{bmatrix}\).
Answer: We need to show that the distributive property of matrix multiplication holds for the given matrices. This means we will calculate \(A(B+C)\) and \(AB+AC\) separately and compare the results.
First, calculate \(B+C\):
\(B+C = \begin{bmatrix} 3 & 1 \\ -1 & 0 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 7 \\ 2 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3+4 & 1+7 \\ -1+2 & 0+1 \\ 4+1 & 2-1 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 1 & 1 \\ 5 & 1 \end{bmatrix}\)
Next, calculate \(A(B+C)\):
\(A(B+C) = \begin{bmatrix} 2 & 0 & -3 \\ 1 & 4 & 5 \end{bmatrix} \begin{bmatrix} 7 & 8 \\ 1 & 1 \\ 5 & 1 \end{bmatrix}\)
\( = \begin{bmatrix} 2(7)+0(1)+(-3)(5) & 2(8)+0(1)+(-3)(1) \\ 1(7)+4(1)+5(5) & 1(8)+4(1)+5(1) \end{bmatrix}\)
\( = \begin{bmatrix} 14+0-15 & 16+0-3 \\ 7+4+25 & 8+4+5 \end{bmatrix}\)
\( = \begin{bmatrix} -1 & 13 \\ 36 & 17 \end{bmatrix}\) (Equation 1)
Now, calculate AB:
\(AB = \begin{bmatrix} 2 & 0 & -3 \\ 1 & 4 & 5 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 0 \\ 4 & 2 \end{bmatrix}\)
\( = \begin{bmatrix} 2(3)+0(-1)+(-3)(4) & 2(1)+0(0)+(-3)(2) \\ 1(3)+4(-1)+5(4) & 1(1)+4(0)+5(2) \end{bmatrix}\)
\( = \begin{bmatrix} 6+0-12 & 2+0-6 \\ 3-4+20 & 1+0+10 \end{bmatrix}\)
\( = \begin{bmatrix} -6 & -4 \\ 19 & 11 \end{bmatrix}\)
Next, calculate AC:
\(AC = \begin{bmatrix} 2 & 0 & -3 \\ 1 & 4 & 5 \end{bmatrix} \begin{bmatrix} 4 & 7 \\ 2 & 1 \\ 1 & -1 \end{bmatrix}\)
\( = \begin{bmatrix} 2(4)+0(2)+(-3)(1) & 2(7)+0(1)+(-3)(-1) \\ 1(4)+4(2)+5(1) & 1(7)+4(1)+5(-1) \end{bmatrix}\)
\( = \begin{bmatrix} 8+0-3 & 14+0+3 \\ 4+8+5 & 7+4-5 \end{bmatrix}\)
\( = \begin{bmatrix} 5 & 17 \\ 17 & 6 \end{bmatrix}\)
Finally, calculate \(AB+AC\):
\(AB+AC = \begin{bmatrix} -6 & -4 \\ 19 & 11 \end{bmatrix} + \begin{bmatrix} 5 & 17 \\ 17 & 6 \end{bmatrix}\)
\( = \begin{bmatrix} -6+5 & -4+17 \\ 19+17 & 11+6 \end{bmatrix}\)
\( = \begin{bmatrix} -1 & 13 \\ 36 & 17 \end{bmatrix}\) (Equation 2)
Comparing Equation 1 and Equation 2, we see that \(A(B+C) = AB+AC\). This verifies the distributive property for matrix multiplication. This property is quite useful for simplifying matrix expressions involving addition and multiplication.
In simple words: First, add matrices B and C, then multiply matrix A by this sum. This is the left side. For the right side, multiply A by B, then multiply A by C, and then add these two results. If both sides give the same final matrix, the property is verified.

🎯 Exam Tip: Matrix multiplication is not commutative (AB \(\neq\) BA) but it is distributive. Keep track of the order of matrices during multiplication to avoid mistakes.

Question 14. Find the matrix A which satisfies the matrix relation \(A \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}\).
Answer: Let the given relation be \(AX = Y\). We need to find matrix A. If X is a \(2 \times 3\) matrix and Y is a \(2 \times 3\) matrix, then A must be a \(2 \times 2\) matrix for the multiplication to work out correctly.
Let \(A = \begin{bmatrix} x & y \\ z & t \end{bmatrix}\).
So, we have:
\( \begin{bmatrix} x & y \\ z & t \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \)
Perform the matrix multiplication on the left side:
\( \begin{bmatrix} x(1)+y(4) & x(2)+y(5) & x(3)+y(6) \\ z(1)+t(4) & z(2)+t(5) & z(3)+t(6) \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \)
Now, equate the corresponding entries to form a system of equations:
For the first row:
1. \(x + 4y = -7\)
2. \(2x + 5y = -8\)
3. \(3x + 6y = -9\) (This simplifies to \(x+2y = -3\))
For the second row:
4. \(z + 4t = 2\)
5. \(2z + 5t = 4\)
6. \(3z + 6t = 6\) (This simplifies to \(z+2t = 2\))
Let's solve the system for x and y using equations 1 and 2.
From equation 1, \(x = -7 - 4y\). Substitute this into equation 2:
\(2(-7 - 4y) + 5y = -8\)
\( -14 - 8y + 5y = -8\)
\( -14 - 3y = -8\)
\( -3y = -8 + 14\)
\( -3y = 6\)
\( y = -2 \)
Now substitute \(y = -2\) back into \(x = -7 - 4y\):
\(x = -7 - 4(-2)\)
\(x = -7 + 8\)
\(x = 1 \)
Next, let's solve the system for z and t using equations 4 and 5.
From equation 4, \(z = 2 - 4t\). Substitute this into equation 5:
\(2(2 - 4t) + 5t = 4\)
\( 4 - 8t + 5t = 4\)
\( 4 - 3t = 4\)
\( -3t = 0\)
\( t = 0 \)
Now substitute \(t = 0\) back into \(z = 2 - 4t\):
\(z = 2 - 4(0)\)
\(z = 2 \)
So, the matrix A is \( \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} \). This method of finding unknown matrix elements by setting up and solving a system of linear equations is a standard approach. The consistent values from checking equations 3 and 6 confirm our result.
In simple words: Imagine matrix A has unknown letters for its numbers. Multiply matrix A by the first given matrix. Then, match each number in your new matrix with the corresponding number in the second given matrix. This will create many small equations. Solve these equations to find all the unknown letters, which will give you the complete matrix A.

🎯 Exam Tip: When equating corresponding entries, always form equations for all positions. This provides redundant equations that can be used to cross-check your values and catch arithmetic errors early.

Question 15. If \( A^{T} = \begin{bmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{bmatrix} \), verify the following
(i) \( (A + B)^T = A^T + B^T = B^T + A^T \)
(ii) \( (A - B)^T = A^T - B^T \)
(iii) \( (B^T)^T = B \)
Answer:
Given:
\( A^{T} = \begin{bmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{bmatrix} \)
To find A, we take the transpose of \( A^T \):
\( (A^T)^T = A \)
\( A = \begin{bmatrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{bmatrix} \)

Given:
\( B = \begin{bmatrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{bmatrix} \)
To find \( B^T \), we take the transpose of B:
\( B^T = \begin{bmatrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{bmatrix} \)

(i) Verify \( (A + B)^T = A^T + B^T = B^T + A^T \)
First, calculate \( A+B \):
\( A + B = \begin{bmatrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 4+2 & -1-1 & 2+1 \\ 5+7 & 0+5 & 3-2 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & -2 & 3 \\ 12 & 5 & 1 \end{bmatrix} \)
Now, find \( (A+B)^T \):
\( (A+B)^T = \begin{bmatrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{bmatrix} \) .....(1)

Next, calculate \( A^T + B^T \):
\( A^T + B^T = \begin{bmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{bmatrix} + \begin{bmatrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 4+2 & 5+7 \\ -1-1 & 0+5 \\ 2+1 & 3-2 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{bmatrix} \) .....(2)

Finally, calculate \( B^T + A^T \):
\( B^T + A^T = \begin{bmatrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{bmatrix} + \begin{bmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2+4 & 7+5 \\ -1-1 & 5+0 \\ 1+2 & -2+3 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{bmatrix} \) .....(3)
From equations (1), (2) and (3), we see that all three results are the same. This proves the first property.

(ii) Verify \( (A - B)^T = A^T - B^T \)
First, calculate \( A-B \):
\( A - B = \begin{bmatrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 4-2 & -1-(-1) & 2-1 \\ 5-7 & 0-5 & 3-(-2) \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 0 & 1 \\ -2 & -5 & 5 \end{bmatrix} \)
Now, find \( (A-B)^T \):
\( (A-B)^T = \begin{bmatrix} 2 & -2 \\ 0 & -5 \\ 1 & 5 \end{bmatrix} \) .....(4)

Next, calculate \( A^T - B^T \):
\( A^T - B^T = \begin{bmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 4-2 & 5-7 \\ -1-(-1) & 0-5 \\ 2-1 & 3-(-2) \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -2 \\ 0 & -5 \\ 1 & 5 \end{bmatrix} \) .....(5)
From equations (4) and (5), we see that both results are the same. This proves the second property.

(iii) Verify \( (B^T)^T = B \)
Given \( B^T = \begin{bmatrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{bmatrix} \)
Now, find the transpose of \( B^T \):
\( (B^T)^T = \begin{bmatrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{bmatrix} \)
This result is exactly the original matrix B. This proves the third property.
In simple words: This question checks if you know how to work with matrix transposes. A transpose means you swap rows and columns. We showed that for addition, the transpose of the sum is the sum of the transposes. For subtraction, the transpose of the difference is the difference of the transposes. Also, taking the transpose twice gets you back to the original matrix.

🎯 Exam Tip: Remember that matrix addition and subtraction are only possible if the matrices have the same dimensions. Transposing a matrix changes its dimensions (rows become columns and vice versa).

Question 16. If A is a 3 x 4 matrix and B is a matrix such that both \( A^{T}B \) and \( BA^{T} \) are defined, what is the order of the matrix B?
Answer:
Given that matrix A is of order \( 3 \times 4 \).
So, its transpose \( A^T \) will be of order \( 4 \times 3 \).

For the product \( A^T B \) to be defined, the number of columns in \( A^T \) must be equal to the number of rows in B.
Since \( A^T \) has 3 columns, B must have 3 rows. So, B is of order \( 3 \times n \).

For the product \( BA^T \) to be defined, the number of columns in B must be equal to the number of rows in \( A^T \).
Since \( A^T \) has 4 rows, B must have 4 columns. So, B is of order \( m \times 4 \).

By combining both conditions, B must have 3 rows and 4 columns.
Therefore, the order of matrix B is \( 3 \times 4 \).
In simple words: If you can multiply a matrix by the transpose of another matrix in two different ways, then the size of the second matrix is fixed. Here, matrix B must be a 3x4 matrix because of how its rows and columns need to match A's transpose for both multiplications to work.

🎯 Exam Tip: Always remember the rule for matrix multiplication: for matrices X and Y, XY is defined only if the number of columns in X equals the number of rows in Y.

Question 17. Express the following matrices as the sum of a symmetric matrix and a skew-symmetric matrix:
(i) \( \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \)
Answer:
(i) Let the given matrix be \( A = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} \).
The transpose of A is \( A^T = \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix} \).

We know that any square matrix A can be expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where \( P = \frac{1}{2}(A + A^T) \) and \( Q = \frac{1}{2}(A - A^T) \).

First, find \( A + A^T \):
\( A + A^T = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix} \)
\( = \begin{bmatrix} 4+4 & -2+3 \\ 3-2 & -5-5 \end{bmatrix} \)
\( = \begin{bmatrix} 8 & 1 \\ 1 & -10 \end{bmatrix} \)

Now, find the symmetric matrix P:
\( P = \frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 8 & 1 \\ 1 & -10 \end{bmatrix} \)
\( P = \begin{bmatrix} 4 & \frac{1}{2} \\ \frac{1}{2} & -5 \end{bmatrix} \)
To check if P is symmetric, we verify \( P^T = P \).
\( P^T = \begin{bmatrix} 4 & \frac{1}{2} \\ \frac{1}{2} & -5 \end{bmatrix}^T = \begin{bmatrix} 4 & \frac{1}{2} \\ \frac{1}{2} & -5 \end{bmatrix} \). Since \( P^T = P \), P is a symmetric matrix.

Next, find \( A - A^T \):
\( A - A^T = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} - \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix} \)
\( = \begin{bmatrix} 4-4 & -2-3 \\ 3-(-2) & -5-(-5) \end{bmatrix} \)
\( = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \)

Now, find the skew-symmetric matrix Q:
\( Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \)
\( Q = \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix} \)
To check if Q is skew-symmetric, we verify \( Q^T = -Q \).
\( Q^T = \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & \frac{5}{2} \\ -\frac{5}{2} & 0 \end{bmatrix} \)
\( -Q = -\begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{2} \\ -\frac{5}{2} & 0 \end{bmatrix} \). Since \( Q^T = -Q \), Q is a skew-symmetric matrix.

Therefore, A can be expressed as the sum of P and Q:
\( A = P + Q = \begin{bmatrix} 4 & \frac{1}{2} \\ \frac{1}{2} & -5 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 4+0 & \frac{1}{2} - \frac{5}{2} \\ \frac{1}{2} + \frac{5}{2} & -5+0 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & -\frac{4}{2} \\ \frac{6}{2} & -5 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} \). This matches the original matrix A.

(ii) Let the given matrix be \( A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \).
The transpose of A is \( A^T = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \).

First, find \( A + A^T \):
\( A + A^T = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 3+3 & 3-2 & -1-4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 & -5+1 & 2+2 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \)

Now, find the symmetric matrix P:
\( P = \frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \)
\( P = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} \)
To check if P is symmetric, we verify \( P^T = P \). This is true.

Next, find \( A - A^T \):
\( A - A^T = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 3-3 & 3-(-2) & -1-(-4) \\ -2-3 & -2-(-2) & 1-(-5) \\ -4-(-1) & -5-1 & 2-2 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} \)

Now, find the skew-symmetric matrix Q:
\( Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} \)
\( Q = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix} \)
To check if Q is skew-symmetric, we verify \( Q^T = -Q \). This is true.

Therefore, A can be expressed as the sum of P and Q:
\( A = P + Q = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 3+0 & \frac{1}{2} + \frac{5}{2} & -\frac{5}{2} + \frac{3}{2} \\ \frac{1}{2} - \frac{5}{2} & -2+0 & -2+3 \\ -\frac{5}{2} - \frac{3}{2} & -2-3 & 2+0 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & \frac{6}{2} & -\frac{2}{2} \\ -\frac{4}{2} & -2 & 1 \\ -\frac{8}{2} & -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \). This matches the original matrix A.
In simple words: Any square matrix can be split into two special types of matrices: one that stays the same when you swap rows and columns (symmetric), and one that becomes its negative when you swap rows and columns (skew-symmetric). We find these two parts by using simple formulas involving the original matrix and its transpose.

🎯 Exam Tip: Remember the formulas \( P = \frac{1}{2}(A + A^T) \) for the symmetric part and \( Q = \frac{1}{2}(A - A^T) \) for the skew-symmetric part. Always verify your symmetric and skew-symmetric matrices by checking their transpose properties.

Question 14. Find the matrix A which satisfies the matrix relation \( A\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \).
Answer:
Let the given equation be \( A X = B \), where \( X = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \).
Matrix X has dimensions \( 2 \times 3 \). Matrix B has dimensions \( 2 \times 3 \).
For \( AX=B \) to be defined, if A is of order \( p \times q \), then X must have q rows. So \( q=2 \).
The resulting matrix B will have p rows and 3 columns. So \( p=2 \).
Thus, A must be a \( 2 \times 2 \) matrix. Let \( A = \begin{bmatrix} x & y \\ z & t \end{bmatrix} \).

Substitute A into the equation:
\( \begin{bmatrix} x & y \\ z & t \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \)

Perform the matrix multiplication on the left side:
\( \begin{bmatrix} (x)(1)+(y)(4) & (x)(2)+(y)(5) & (x)(3)+(y)(6) \\ (z)(1)+(t)(4) & (z)(2)+(t)(5) & (z)(3)+(t)(6) \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \)
\( \begin{bmatrix} x+4y & 2x+5y & 3x+6y \\ z+4t & 2z+5t & 3z+6t \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \)

Equate corresponding entries to form a system of equations:
From the first row:
1) \( x+4y = -7 \)
2) \( 2x+5y = -8 \)
3) \( 3x+6y = -9 \)

From the second row:
4) \( z+4t = 2 \)
5) \( 2z+5t = 4 \)
6) \( 3z+6t = 6 \)

Let's solve for x and y using equations (1) and (2):
Multiply equation (1) by 2: \( 2(x+4y) = 2(-7) \implies 2x+8y = -14 \)
Subtract equation (2) from this new equation:
\( (2x+8y) - (2x+5y) = -14 - (-8) \)
\( 3y = -6 \)
\( y = \frac{-6}{3} \)
\( y = -2 \)

Substitute \( y=-2 \) into equation (1):
\( x+4(-2) = -7 \)
\( x-8 = -7 \)
\( x = -7+8 \)
\( x = 1 \)

Now, let's solve for z and t using equations (4) and (5):
Multiply equation (4) by 2: \( 2(z+4t) = 2(2) \implies 2z+8t = 4 \)
Subtract equation (5) from this new equation:
\( (2z+8t) - (2z+5t) = 4 - 4 \)
\( 3t = 0 \)
\( t = 0 \)

Substitute \( t=0 \) into equation (4):
\( z+4(0) = 2 \)
\( z+0 = 2 \)
\( z = 2 \)

So, the values are \( x=1, y=-2, z=2, t=0 \).
Therefore, the required matrix A is \( \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} \).
In simple words: We are looking for a hidden matrix A. We know its size must be 2x2. By multiplying the hidden matrix A with the given matrix and then comparing the result, we get several small math problems. Solving these problems helps us find the numbers inside matrix A one by one.

🎯 Exam Tip: When finding an unknown matrix in a product, first determine its dimensions by looking at the dimensions of the known matrices. This helps set up the unknown matrix with variables correctly.

Question 19. If \( A = \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y \end{matrix} \right] \) is a matrix such that \( AA^T = 9I \), find the values of x and y.
Answer:
Given matrix \( A = \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{matrix} \right] \).
First, we find the transpose of A:
\( A^T = \left[ \begin{matrix} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{matrix} \right] \)
Now, we multiply A by \( A^T \):
\( AA^T = \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{matrix} \right] \)
\( = \left[ \begin{matrix} (1)(1)+(2)(2)+(2)(2) & (1)(2)+(2)(1)+(2)(-2) & (1)(x)+(2)(2)+(2)(y) \\ (2)(1)+(1)(2)+(-2)(2) & (2)(2)+(1)(1)+(-2)(-2) & (2)(x)+(1)(2)+(-2)(y) \\ (x)(1)+(2)(2)+(y)(2) & (x)(2)+(2)(1)+(y)(-2) & (x)(x)+(2)(2)+(y)(y) \end{matrix} \right] \)
\( = \left[ \begin{matrix} 1+4+4 & 2+2-4 & x+4+2y \\ 2+2-4 & 4+1+4 & 2x+2-2y \\ x+4+2y & 2x+2-2y & x^2+4+y^2 \end{matrix} \right] \)
\( = \left[ \begin{matrix} 9 & 0 & x+4+2y \\ 0 & 9 & 2x+2-2y \\ x+4+2y & 2x+2-2y & x^2+4+y^2 \end{matrix} \right] \)
We are given that \( AA^T = 9I \), where I is the identity matrix of order 3:
\( 9I = 9 \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] = \left[ \begin{matrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{matrix} \right] \)
Now, we equate the corresponding entries of \( AA^T \) and \( 9I \):
\( x+4+2y = 0 \) (1)
\( 2x+2-2y = 0 \) (2)
\( x^2+4+y^2 = 9 \) (3)
From equation (2), we can simplify it by dividing by 2:
\( x+1-y = 0 \)
\( x-y = -1 \) (4)
From equation (1):
\( x+2y = -4 \) (5)
Now, subtract equation (4) from equation (5):
\( (x+2y) - (x-y) = -4 - (-1) \)
\( x+2y-x+y = -4+1 \)
\( 3y = -3 \)
\( y = -1 \)
Substitute \( y = -1 \) into equation (4):
\( x - (-1) = -1 \)
\( x+1 = -1 \)
\( x = -2 \)
Finally, we check these values with equation (3):
\( x^2+4+y^2 = (-2)^2+4+(-1)^2 = 4+4+1 = 9 \). This matches the required value.
So, the values of x and y are -2 and -1 respectively. The identity matrix plays a crucial role in verifying the properties of square matrices.
In simple words: We first found the transpose of matrix A and multiplied it by A. Then we set this new matrix equal to 9 times the identity matrix. By comparing the elements, we formed equations and solved them to find that x is -2 and y is -1.

🎯 Exam Tip: When equating matrices, make sure to set up all possible independent equations from the corresponding entries. Always double-check your final values by substituting them back into the original condition.

Question 20. (i) For what value of x, the matrix A is skew-symmetric
\( A = \left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & x^3 \\ 2 & -3 & 0 \end{matrix} \right] \)
Answer:
A matrix A is skew-symmetric if its transpose is equal to the negative of the matrix, i.e., \( A^T = -A \).
Given matrix \( A = \left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & x^3 \\ 2 & -3 & 0 \end{matrix} \right] \)
First, find the transpose of A:
\( A^T = \left[ \begin{matrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & x^3 & 0 \end{matrix} \right] \)
Next, find the negative of A:
\( -A = \left[ \begin{matrix} 0 & -1 & 2 \\ 1 & 0 & -x^3 \\ -2 & 3 & 0 \end{matrix} \right] \)
For A to be skew-symmetric, \( A^T = -A \). Equating the corresponding elements:
Comparing the element at row 2, column 3 (second row, third column):
\( (A^T)_{23} = -3 \)
\( (-A)_{23} = -x^3 \)
So, \( -3 = -x^3 \)
\( \implies x^3 = 3 \)
This means \( x = \sqrt[3]{3} \).
This value ensures the matrix is skew-symmetric. Remember, the diagonal elements of a skew-symmetric matrix are always zero.
In simple words: For a matrix to be skew-symmetric, its mirror image across the diagonal must be the negative of the original. By comparing the elements of the transposed matrix and the negative of the original matrix, we found that x cubed must be equal to 3.

🎯 Exam Tip: Always remember that the diagonal elements of a skew-symmetric matrix must be zero, and \( a_{ij} = -a_{ji} \) for all i and j. Use these conditions to quickly check your answer.

Question 20. (ii) If \( \left[ \begin{matrix} 0 & p & 3 \\ 2 & q^2 & -1 \\ r & 1 & 0 \end{matrix} \right] \) is skew-symmetric find the values of p, q and r.
Answer:
Let the given matrix be \( A = \left[ \begin{matrix} 0 & p & 3 \\ 2 & q^2 & -1 \\ r & 1 & 0 \end{matrix} \right] \).
For A to be skew-symmetric, \( A^T = -A \).
First, find the transpose of A:
\( A^T = \left[ \begin{matrix} 0 & 2 & r \\ p & q^2 & 1 \\ 3 & -1 & 0 \end{matrix} \right] \)
Next, find the negative of A:
\( -A = \left[ \begin{matrix} 0 & -p & -3 \\ -2 & -q^2 & 1 \\ -r & -1 & 0 \end{matrix} \right] \)
Now, equate \( A^T \) and \( -A \):
\( \left[ \begin{matrix} 0 & 2 & r \\ p & q^2 & 1 \\ 3 & -1 & 0 \end{matrix} \right] = \left[ \begin{matrix} 0 & -p & -3 \\ -2 & -q^2 & 1 \\ -r & -1 & 0 \end{matrix} \right] \)
Equating the corresponding entries:
Comparing \( a_{12} \) entries: \( 2 = -p \implies p = -2 \)
Comparing \( a_{13} \) entries: \( r = -3 \)
Comparing \( a_{21} \) entries: \( p = -2 \) (This matches the value of p found earlier)
Comparing \( a_{22} \) entries: \( q^2 = -q^2 \)
\( \implies 2q^2 = 0 \)
\( \implies q^2 = 0 \)
\( \implies q = 0 \)
Comparing \( a_{31} \) entries: \( 3 = -r \implies r = -3 \) (This matches the value of r found earlier)
Thus, the values are \( p = -2 \), \( q = 0 \), and \( r = -3 \). This condition is fundamental for understanding matrix types.
In simple words: We used the rule that a skew-symmetric matrix, when transposed, is the same as its negative. By comparing the elements of the given matrix's transpose with its negative, we found the values for p, q, and r.

🎯 Exam Tip: Remember that for a skew-symmetric matrix, \( a_{ij} = -a_{ji} \) for all elements, and the main diagonal elements \( a_{ii} \) must always be zero. These properties are key to solving such problems efficiently.

Question 21. Construct the matrix \( A = [a_{ij}]_{3 \times 3} \), where \( a_{ij} = i-j \). State whether A is symmetric or skew - symmetric.
Answer:
To construct the \( 3 \times 3 \) matrix A with elements \( a_{ij} = i-j \), we calculate each element:
\( a_{11} = 1-1 = 0 \)
\( a_{12} = 1-2 = -1 \)
\( a_{13} = 1-3 = -2 \)
\( a_{21} = 2-1 = 1 \)
\( a_{22} = 2-2 = 0 \)
\( a_{23} = 2-3 = -1 \)
\( a_{31} = 3-1 = 2 \)
\( a_{32} = 3-2 = 1 \)
\( a_{33} = 3-3 = 0 \)
So, the matrix A is:
\( A = \left[ \begin{matrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] \)
Now, we find the transpose of A, denoted as \( A^T \):
\( A^T = \left[ \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{matrix} \right] \)
To check if A is symmetric, we compare A with \( A^T \). A is symmetric if \( A = A^T \). Clearly, A is not equal to \( A^T \).
To check if A is skew-symmetric, we compare \( A^T \) with \( -A \). A is skew-symmetric if \( A^T = -A \).
First, find \( -A \):
\( -A = \left[ \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{matrix} \right] \)
Since \( A^T = -A \), the matrix A is skew-symmetric. This property is crucial in advanced linear algebra.
In simple words: We built the matrix A using the rule \( i-j \) for each spot. Then we flipped the rows and columns to get the transpose, \( A^T \). When we compared \( A^T \) with \( -A \) (which is A with all signs flipped), they matched, meaning the matrix is skew-symmetric.

🎯 Exam Tip: For problems involving matrix construction from a formula, calculate each element carefully. To determine if a matrix is symmetric or skew-symmetric, always calculate its transpose first and then compare it with the original matrix and its negative.

Question 22. Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.
Answer:
Given that A and B are symmetric matrices. This means:
\( A = A^T \) (1)
\( B = B^T \) (2)
We need to prove that AB = BA if and only if AB is a symmetric matrix.

**Part 1: Assume AB = BA, prove that AB is symmetric.**
If AB = BA, we want to show that \( (AB)^T = AB \).
Let's find the transpose of AB:
\( (AB)^T = B^T A^T \) (Using the property \( (XY)^T = Y^T X^T \))
Since A and B are symmetric (from (1) and (2)), we can replace \( B^T \) with B and \( A^T \) with A:
\( (AB)^T = BA \)
We assumed that AB = BA, so we can substitute AB for BA:
\( (AB)^T = AB \)
Therefore, if AB = BA, then AB is a symmetric matrix.

**Part 2: Assume AB is symmetric, prove that AB = BA.**
If AB is symmetric, then by definition:
\( (AB)^T = AB \) (3)
We also know that for any matrices X and Y:
\( (AB)^T = B^T A^T \) (Using the property \( (XY)^T = Y^T X^T \))
Since A and B are symmetric (from (1) and (2)), we can replace \( B^T \) with B and \( A^T \) with A:
\( (AB)^T = BA \)
Now, from equation (3), we have \( (AB)^T = AB \). So, we can write:
\( AB = BA \)
Therefore, if AB is a symmetric matrix, then AB = BA.
Since both parts of the "if and only if" condition have been proven, the statement is true. Matrix properties are fundamental to linear algebra operations.
In simple words: We showed that if matrix A multiplied by B is the same as B multiplied by A, then their product (AB) is a symmetric matrix. We also showed that if the product AB is symmetric, then A multiplied by B is the same as B multiplied by A. Both directions are true, proving the statement.

🎯 Exam Tip: For "if and only if" proofs, you must prove both directions: (1) if P then Q, and (2) if Q then P. Remember the transpose properties: \( (A+B)^T = A^T+B^T \) and \( (AB)^T = B^T A^T \).

Question 23. If A and B are symmetric matrices of the same order, prove that
(i) AB + BA is a symmetric matrix
(ii) AB – BA is a skew-symmetric matrix.
Answer:
Given that A and B are symmetric matrices of the same order.
This means: \( A = A^T \) and \( B = B^T \).

**(i) To prove AB + BA is a symmetric matrix:**
A matrix X is symmetric if \( X^T = X \). So we need to show \( (AB+BA)^T = AB+BA \).
Let's take the transpose of the sum (AB + BA):
\( (AB+BA)^T = (AB)^T + (BA)^T \) (Using the property \( (X+Y)^T = X^T+Y^T \))
Now, apply the property for the transpose of a product \( (XY)^T = Y^T X^T \):
\( (AB)^T = B^T A^T \)
\( (BA)^T = A^T B^T \)
So, \( (AB+BA)^T = B^T A^T + A^T B^T \)
Since A and B are given as symmetric matrices, we know \( A^T = A \) and \( B^T = B \). Substitute these into the expression:
\( (AB+BA)^T = BA + AB \)
Since matrix addition is commutative (AB + BA = BA + AB), we can write:
\( (AB+BA)^T = AB + BA \)
Therefore, AB + BA is a symmetric matrix. This demonstrates how matrix properties interact.

**(ii) To prove AB – BA is a skew-symmetric matrix:**
A matrix X is skew-symmetric if \( X^T = -X \). So we need to show \( (AB-BA)^T = -(AB-BA) \).
Let's take the transpose of the difference (AB - BA):
\( (AB-BA)^T = (AB)^T - (BA)^T \) (Using the property \( (X-Y)^T = X^T-Y^T \))
Now, apply the property for the transpose of a product \( (XY)^T = Y^T X^T \):
\( (AB)^T = B^T A^T \)
\( (BA)^T = A^T B^T \)
So, \( (AB-BA)^T = B^T A^T - A^T B^T \)
Since A and B are given as symmetric matrices, we know \( A^T = A \) and \( B^T = B \). Substitute these into the expression:
\( (AB-BA)^T = BA - AB \)
We want to show this is equal to \( -(AB-BA) \). Let's expand \( -(AB-BA) \):
\( -(AB-BA) = -AB + BA = BA - AB \)
Since \( (AB-BA)^T = BA - AB \) and \( -(AB-BA) = BA - AB \), we have:
\( (AB-BA)^T = -(AB-BA) \)
Therefore, AB - BA is a skew-symmetric matrix. These types of proofs are common in matrix theory.
In simple words: For part (i), we found that the transpose of (AB + BA) is equal to (BA + AB), which is the same as (AB + BA). This means it's a symmetric matrix. For part (ii), we found that the transpose of (AB - BA) is (BA - AB), which is the negative of (AB - BA). This means it's a skew-symmetric matrix.

🎯 Exam Tip: Clearly state the definitions of symmetric and skew-symmetric matrices at the beginning of your proof. Carefully apply the properties of transposes for sums and products, replacing \( A^T \) with A and \( B^T \) with B as symmetric matrices allow.

Question 24. A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds. The pack contains 100 gm of cashew nuts, 100 gm of raisins, and 50 gm of almonds. Pack – II contains 200 gm of cashew nuts, 100 gm of raisins, and 100 gm of almonds. Pack -III contains 250 gm of cashew nuts, 250 gm of raisins, and 150 gm of almonds. The cost of 50 gm of cashew nuts is Rs. 50, 50 gm of raisins is Rs. 10, and 50 gm of almonds is Rs. 60. What is the cost of each gift pack?
Answer:
First, let's define a "unit" as 50 gm for easier calculation, since all costs are given per 50 gm.
Cost per unit (50 gm):
Cashew nuts: Rs. 50
Raisins: Rs. 10
Almonds: Rs. 60

Now, let's represent the composition of each gift pack in terms of these units:

**Pack I:**
Cashew nuts: 100 gm = 2 units
Raisins: 100 gm = 2 units
Almonds: 50 gm = 1 unit

**Pack II:**
Cashew nuts: 200 gm = 4 units
Raisins: 100 gm = 2 units
Almonds: 100 gm = 2 units

**Pack III:**
Cashew nuts: 250 gm = 5 units
Raisins: 250 gm = 5 units
Almonds: 150 gm = 3 units

We can represent the quantities of nuts in each pack as a matrix A, where rows represent nut types (cashew, raisins, almonds) and columns represent the packs (I, II, III):
\( A = \left[ \begin{matrix} 2 & 4 & 5 \\ 2 & 2 & 5 \\ 1 & 2 & 3 \end{matrix} \right] \)

The cost per unit for each type of nut can be represented as a row matrix B:
\( B = \left[ \begin{matrix} 50 & 10 & 60 \end{matrix} \right] \)

To find the total cost of each gift pack, we multiply the cost matrix B by the quantity matrix A (B will represent the cost for each item in the pack, and A shows how many units of each item are in each pack):
Cost of gift packs = \( BA \)
\( BA = \left[ \begin{matrix} 50 & 10 & 60 \end{matrix} \right] \left[ \begin{matrix} 2 & 4 & 5 \\ 2 & 2 & 5 \\ 1 & 2 & 3 \end{matrix} \right] \)
\( BA = \left[ \begin{matrix} (50)(2)+(10)(2)+(60)(1) & (50)(4)+(10)(2)+(60)(2) & (50)(5)+(10)(5)+(60)(3) \end{matrix} \right] \)
\( BA = \left[ \begin{matrix} 100+20+60 & 200+20+120 & 250+50+180 \end{matrix} \right] \)
\( BA = \left[ \begin{matrix} 180 & 340 & 480 \end{matrix} \right] \)

Therefore, the cost of each gift pack is:
Cost of Gift Pack I = Rs. 180
Cost of Gift Pack II = Rs. 340
Cost of Gift Pack III = Rs. 480
Matrix multiplication helps in organizing and calculating costs for multiple products efficiently.
In simple words: We first figured out the cost for a small unit (50 grams) of each item. Then, we made two tables (matrices): one showing how many units of each item are in each gift pack, and another showing the cost per unit. By multiplying these tables, we found the total cost for each of the three gift packs.

🎯 Exam Tip: Always ensure your matrices are set up correctly for multiplication based on their dimensions and what each row/column represents. If costs are given per different units (e.g., per 50gm and per 100gm), normalize them to a common unit before forming the cost matrix.