Samacheer Kalvi Class 11 Maths Solutions Chapter 7 Matrices and Determinants Exercise 7.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1. If \( a_{ij} = (3i – 2j) \) and \( A = [a_{ij}]_{3 \times 2} \) is
(a) \[ \begin{bmatrix} 1 & -2 \\ 2 & 1 \\ 2 & 1 \end{bmatrix} \]
(b) \[ \begin{bmatrix} 1 & -1 \\ 2 & 2 \\ 2 & 1 \end{bmatrix} \]
(c) \[ \begin{bmatrix} 1 & -1 \\ -2 & -2 \\ 1 & 2 \end{bmatrix} \]
(d) \[ \begin{bmatrix} -1 & -1 \\ 2 & -2 \\ 1 & 2 \end{bmatrix} \]
Answer: (b) \[ \begin{bmatrix} 1 & -1 \\ 2 & 2 \\ 2 & 1 \end{bmatrix} \]
In simple words: To find the matrix A, we use the formula \( a_{ij} \) to calculate each element. The row number is \( i \) and the column number is \( j \). We substitute these numbers into the formula to get the value for each spot in the matrix.

🎯 Exam Tip: Always remember that the first subscript \( i \) indicates the row and the second subscript \( j \) indicates the column when constructing a matrix.

Question 2. What must be the matrix X, if \( 2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} \)?
(a) \[ \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \]
(b) \[ \begin{bmatrix} 1 & -3 \\ -2 & 1 \end{bmatrix} \]
(c) \[ \begin{bmatrix} 4 & -2 \\ 2 & -6 \end{bmatrix} \]
(d) \[ \begin{bmatrix} 4 & -2 \\ 4 & -2 \end{bmatrix} \]
Answer: (a) \[ \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \]
\[ 2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} \] Adding \( - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) on both sides:
\[ 2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \] \[ 2X + \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 3-1 & 8-2 \\ 7-3 & 2-4 \end{bmatrix} \] \[ 2X = \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix} \] Now, multiply both sides by \( \frac{1}{2} \):
\[ X = \frac{1}{2} \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix} \] \[ X = \begin{bmatrix} \frac{2}{2} & \frac{6}{2} \\ \frac{4}{2} & \frac{-2}{2} \end{bmatrix} \] \[ X = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \]
In simple words: To find matrix X, first subtract the known matrix from both sides of the equation. Then, divide all the numbers in the resulting matrix by 2. This isolates X and gives its final values.

🎯 Exam Tip: Remember that matrix addition/subtraction and scalar multiplication are performed element-wise. Don't mix up matrix operations with scalar arithmetic.

Question 3. Which one of the following is not true about the matrix \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)?
(a) a scalar matrix
(b) a diagonal matrix
(c) an upper triangular matrix
(d) a lower triangular matrix
Answer: (a) a scalar matrix
Explanation:
A scalar matrix is a diagonal matrix where all the elements on the main diagonal are equal. The given matrix is \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] 1. This is a diagonal matrix because all non-diagonal elements are zero. 2. It is an upper triangular matrix because all elements below the main diagonal are zero. 3. It is a lower triangular matrix because all elements above the main diagonal are zero. 4. However, it is not a scalar matrix because the main diagonal elements (1, 0, 0) are not all equal. For example, the element \( a_{11} = 1 \) while \( a_{22} = 0 \). Therefore, the statement "a scalar matrix" is not true for this matrix.
In simple words: A scalar matrix needs all the numbers on its main diagonal to be the same. In this matrix, the numbers on the diagonal are 1, 0, and 0, which are not all equal, so it is not a scalar matrix.

🎯 Exam Tip: To identify matrix types, focus on the positions of zero and non-zero elements, especially along and relative to the main diagonal.

Question 4. If A and B are two matrices such that A + B and AB are both defined, then
(a) A and B are two matrices not necessarily of same order
(b) A and B are square matrices of same order
(c) Number of columns of A is equal to the number of rows of B
(d) A = B
Answer: (b) A and B are square matrices of same order
Explanation:
If A and B are two matrices: 1. For A + B to be defined, matrices A and B must have the same order (i.e., same number of rows and same number of columns). 2. For AB to be defined, the number of columns in A must be equal to the number of rows in B. If both A + B and AB are defined, let's say A is an \( m \times n \) matrix and B is a \( p \times q \) matrix. From condition (1), A and B must be of the same order, so \( m = p \) and \( n = q \). From condition (2), the number of columns in A must equal the number of rows in B, so \( n = p \). Combining these, we get \( m = p = n = q \). This means both A and B must be square matrices of the same order. For instance, if A is \( 2 \times 3 \) and B is \( 2 \times 3 \), A+B is defined. But for AB to be defined, the column of A (3) must equal the row of B (2), which is false. Therefore, they must be square matrices of the same order.
In simple words: If you can add two matrices, they must be the same size. If you can multiply them, the number of columns in the first must match the number of rows in the second. For both to be possible, both matrices must be square and also the same size as each other.

🎯 Exam Tip: Remember the fundamental rules for matrix addition (same dimensions) and multiplication (inner dimensions match). These rules are key to solving questions about matrix definitions.

Question 5. If \( A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix} \), then for what value of \( \lambda \), \( A^2 = 0 \)?
(a) 0
(b) \( \pm 1 \)
(c) - 1
(d) 1
Answer: (b) \( \pm 1 \)
Explanation:
Given matrix \( A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix} \).
To find \( A^2 \), we multiply A by A:
\[ A^2 = A \cdot A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix} \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix} \] \[ A^2 = \begin{bmatrix} (\lambda)(\lambda) + (1)(-1) & (\lambda)(1) + (1)(-\lambda) \\ (-1)(\lambda) + (-\lambda)(-1) & (-1)(1) + (-\lambda)(-\lambda) \end{bmatrix} \] \[ A^2 = \begin{bmatrix} \lambda^2 - 1 & \lambda - \lambda \\ -\lambda + \lambda & -1 + \lambda^2 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{bmatrix} \] We are given that \( A^2 = 0 \), where \( 0 \) is the zero matrix \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
So, we set the elements of \( A^2 \) equal to the elements of the zero matrix:
\[ \begin{bmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \] This means that \( \lambda^2 - 1 = 0 \).
Solving for \( \lambda \):
\( \lambda^2 = 1 \)
\( \implies \) \( \lambda = \pm \sqrt{1} \)
\( \implies \) \( \lambda = \pm 1 \).
So, the values of \( \lambda \) for which \( A^2 = 0 \) are \( 1 \) and \( -1 \).
In simple words: We multiply the matrix A by itself to get \( A^2 \). Then, we set all the numbers in \( A^2 \) to zero because it's a zero matrix. By solving the simple equation \( \lambda^2 - 1 = 0 \), we find that \( \lambda \) can be either 1 or -1.

🎯 Exam Tip: Remember that a matrix is equal to a zero matrix only if all its elements are zero. When equating matrix elements, ensure each corresponding element is set to zero.

Question 6. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \) and \( (A + B)^2 = A^2 + B^2 \), then the values of a and b are
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, b = 4
(d) a = 2, b = 4
Answer: (b) a = 1, b = 4
Explanation:
Given matrices \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \).
We are also given the condition \( (A + B)^2 = A^2 + B^2 \). First, calculate \( A^2 \):
\[ A^2 = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(-1)(2) & (1)(-1)+(-1)(-1) \\ (2)(1)+(-1)(2) & (2)(-1)+(-1)(-1) \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 1-2 & -1+1 \\ 2-2 & -2+1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \] Next, calculate \( B^2 \):
\[ B^2 = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = \begin{bmatrix} (a)(a)+(1)(b) & (a)(1)+(1)(-1) \\ (b)(a)+(-1)(b) & (b)(1)+(-1)(-1) \end{bmatrix} \] \[ B^2 = \begin{bmatrix} a^2+b & a-1 \\ ab-b & b+1 \end{bmatrix} \] Now, calculate \( A^2 + B^2 \):
\[ A^2 + B^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} a^2+b & a-1 \\ ab-b & b+1 \end{bmatrix} \] \[ A^2 + B^2 = \begin{bmatrix} -1+a^2+b & 0+a-1 \\ 0+ab-b & -1+b+1 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \\ ab-b & b \end{bmatrix} \quad \ldots (1) \] Now, calculate \( A + B \):
\[ A + B = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = \begin{bmatrix} 1+a & -1+1 \\ 2+b & -1-1 \end{bmatrix} \] \[ A + B = \begin{bmatrix} a+1 & 0 \\ 2+b & -2 \end{bmatrix} \] Next, calculate \( (A+B)^2 \):
\[ (A+B)^2 = \begin{bmatrix} a+1 & 0 \\ 2+b & -2 \end{bmatrix} \begin{bmatrix} a+1 & 0 \\ 2+b & -2 \end{bmatrix} \] \[ (A+B)^2 = \begin{bmatrix} (a+1)(a+1)+0(2+b) & (a+1)(0)+0(-2) \\ (2+b)(a+1)+(-2)(2+b) & (2+b)(0)+(-2)(-2) \end{bmatrix} \] \[ (A+B)^2 = \begin{bmatrix} (a+1)^2 & 0 \\ (2+b)(a+1-2) & 4 \end{bmatrix} \] \[ (A+B)^2 = \begin{bmatrix} (a+1)^2 & 0 \\ (2+b)(a-1) & 4 \end{bmatrix} \quad \ldots (2) \] We are given \( (A+B)^2 = A^2+B^2 \). So, equating matrices (1) and (2):
\[ \begin{bmatrix} (a+1)^2 & 0 \\ (2+b)(a-1) & 4 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \\ ab-b & b \end{bmatrix} \] Now, equate the corresponding elements:
1. From the (1, 2) element: \( 0 = a-1 \implies a=1 \).
2. From the (2, 2) element: \( 4 = b \). We can check these values using other elements, although not strictly required if we are confident.
Using \( a=1 \) and \( b=4 \):
\( (a+1)^2 = (1+1)^2 = 2^2 = 4 \).
\( a^2+b-1 = 1^2+4-1 = 1+4-1 = 4 \). (Matches)
\( (2+b)(a-1) = (2+4)(1-1) = 6 \times 0 = 0 \).
\( ab-b = (1)(4)-4 = 4-4 = 0 \). (Matches)
The values \( a=1 \) and \( b=4 \) are consistent with the equation.
In simple words: We first calculate \( A^2 \), \( B^2 \), and then their sum. Then, we calculate \( A+B \) and its square. We set these two results equal to each other and compare the elements. This helps us find the values of 'a' and 'b'. The process involves careful matrix multiplication and equating corresponding elements.

🎯 Exam Tip: Be careful with matrix multiplication, as it is not commutative (\( AB \ne BA \)). When equating matrices, ensure every corresponding element is equal to avoid errors.

Question 7. If \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \) the equation \( AA^T = 9I \), where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to
(a) (2, -1)
(b) (- 2, 1)
(c) (2, 1)
(d) (- 2, – 1)
Answer: (d) (- 2, – 1)
Explanation:
Given matrix \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \).
First, find the transpose of A, \( A^T \):
\[ A^T = \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} \] Next, calculate \( AA^T \):
\[ AA^T = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} \] \[ AA^T = \begin{bmatrix} (1)(1)+(2)(2)+(2)(2) & (1)(2)+(2)(1)+(2)(-2) & (1)(a)+(2)(2)+(2)(b) \\ (2)(1)+(1)(2)+(-2)(2) & (2)(2)+(1)(1)+(-2)(-2) & (2)(a)+(1)(2)+(-2)(b) \\ (a)(1)+(2)(2)+(b)(2) & (a)(2)+(2)(1)+(b)(-2) & (a)(a)+(2)(2)+(b)(b) \end{bmatrix} \] \[ AA^T = \begin{bmatrix} 1+4+4 & 2+2-4 & a+4+2b \\ 2+2-4 & 4+1+4 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} \] \[ AA^T = \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+b^2+4 \end{bmatrix} \] We are given that \( AA^T = 9I \), where \( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
So, \( 9I = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \).
Equating \( AA^T \) with \( 9I \):
\[ \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+b^2+4 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \] Equating the corresponding entries, we get the following equations:
1. \( a+4+2b = 0 \quad \ldots (1) \)
2. \( 2a+2-2b = 0 \quad \ldots (2) \)
3. \( a^2+b^2+4 = 9 \quad \ldots (3) \) From equation (2), divide by 2: \( a+1-b = 0 \implies a = b-1 \).
Substitute \( a = b-1 \) into equation (1):
\( (b-1)+4+2b = 0 \)
\( \implies \) \( 3b+3 = 0 \)
\( \implies \) \( 3b = -3 \)
\( \implies \) \( b = -1 \).
Now substitute \( b = -1 \) back into \( a = b-1 \):
\( a = -1-1 \)
\( \implies \) \( a = -2 \).
So, the ordered pair is \( (a, b) = (-2, -1) \).
Let's verify these values with equation (3):
\( (-2)^2 + (-1)^2 + 4 = 4 + 1 + 4 = 9 \). This matches.
In simple words: First, we find the transpose of matrix A. Then, we multiply A by its transpose to get \( AA^T \). We set this result equal to 9 times the identity matrix. By comparing the elements of these matrices, we get equations that we can solve to find the values for 'a' and 'b'. The ordered pair is \( (-2, -1) \).

🎯 Exam Tip: When working with matrix transposes and identities, remember that \( (X^T)^T = X \) and \( I \) is a matrix with 1s on the main diagonal and 0s elsewhere. Be meticulous with algebraic substitutions to avoid errors.

Question 8. If A is a square matrix, then which of the following is not symmetric?
(a) \( A + A^T \)
(b) \( AA^T \)
(c) \( A^T A \)
(d) \( A - A^T \)
Answer: (d) \( A - A^T \)
Explanation:
A matrix X is symmetric if \( X^T = X \). A matrix X is skew-symmetric if \( X^T = -X \). Let's check each option:
(a) For \( A + A^T \):
\( (A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T \).
Since \( (A + A^T)^T = A + A^T \), \( A + A^T \) is symmetric.
(b) For \( AA^T \):
\( (AA^T)^T = (A^T)^T A^T = A A^T \).
Since \( (AA^T)^T = AA^T \), \( AA^T \) is symmetric.
(c) For \( A^T A \):
\( (A^T A)^T = A^T (A^T)^T = A^T A \).
Since \( (A^T A)^T = A^T A \), \( A^T A \) is symmetric.
(d) For \( A - A^T \):
\( (A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T) \).
Since \( (A - A^T)^T = -(A - A^T) \), \( A - A^T \) is skew-symmetric, not symmetric. Therefore, this is the correct answer.
In simple words: A symmetric matrix stays the same when you flip its rows and columns. We check each option by taking its transpose. If the transpose is the same as the original, it's symmetric. For \( A - A^T \), the transpose is the negative of the original, meaning it's skew-symmetric, not symmetric.

🎯 Exam Tip: Remember the properties of transposes: \( (X+Y)^T = X^T+Y^T \), \( (XY)^T = Y^T X^T \), and \( (X^T)^T = X \). These are essential for proving symmetry or skew-symmetry.

Question 9. If A and B are symmetric matrices of order n, where \( (A \ne B) \), then
(a) A + B is skew – symmetric
(b) A + B is symmetric
(c) A + B is a diagonal matrix
(d) A + B is a zero matrix
Answer: (b) A + B is symmetric
Explanation:
Given that A and B are symmetric matrices of order n.
This means \( A^T = A \) and \( B^T = B \).
To check if \( A+B \) is symmetric, we take its transpose:
\( (A+B)^T = A^T + B^T \)
Since A and B are symmetric, we can substitute \( A^T = A \) and \( B^T = B \):
\( (A+B)^T = A + B \).
Since \( (A+B)^T = A+B \), the matrix \( A+B \) is symmetric.
The other options are generally not true for any two symmetric matrices. For example, \( A+B \) is not necessarily a diagonal matrix or a zero matrix. Also, it is not skew-symmetric because its transpose is \( A+B \), not \( -(A+B) \).
In simple words: If both matrices A and B are symmetric (meaning they stay the same when flipped), then their sum A + B will also be symmetric. This is because flipping the sum is the same as flipping each matrix and then adding them, which gives the original sum back.

🎯 Exam Tip: The sum of two symmetric matrices is always symmetric. This is a standard property you should remember for matrix theory questions.

Question 10. If \( A = \begin{bmatrix} a & x \\ y & a \end{bmatrix} \) and if xy = 1, then det \( (AA^T) \) is equal to
(a) \( (a – 1)^2 \)
(b) \( (a^2 + 1)^2 \)
(c) \( a^2 – 1 \)
(d) \( (a^2 – 1)^2 \)
Answer: (d) \( (a^2 – 1)^2 \)
Explanation:
Given matrix \( A = \begin{bmatrix} a & x \\ y & a \end{bmatrix} \) and \( xy = 1 \).
First, find the transpose of A, \( A^T \):
\[ A^T = \begin{bmatrix} a & y \\ x & a \end{bmatrix} \] Next, calculate \( AA^T \):
\[ AA^T = \begin{bmatrix} a & x \\ y & a \end{bmatrix} \begin{bmatrix} a & y \\ x & a \end{bmatrix} \] \[ AA^T = \begin{bmatrix} (a)(a)+(x)(x) & (a)(y)+(x)(a) \\ (y)(a)+(a)(x) & (y)(y)+(a)(a) \end{bmatrix} \] \[ AA^T = \begin{bmatrix} a^2+x^2 & ay+ax \\ ay+ax & y^2+a^2 \end{bmatrix} \] Now, find the determinant of \( AA^T \):
\( \text{det}(AA^T) = (a^2+x^2)(y^2+a^2) - (ay+ax)(ay+ax) \)
\( \implies \) \( \text{det}(AA^T) = (a^2+x^2)(y^2+a^2) - (ay+ax)^2 \)
\( \implies \) \( \text{det}(AA^T) = (a^2y^2 + a^4 + x^2y^2 + a^2x^2) - (a^2y^2 + 2a^2xy + a^2x^2) \)
\( \implies \) \( \text{det}(AA^T) = a^2y^2 + a^4 + x^2y^2 + a^2x^2 - a^2y^2 - 2a^2xy - a^2x^2 \)
\( \implies \) \( \text{det}(AA^T) = a^4 + x^2y^2 - 2a^2xy \)
We are given \( xy=1 \). Substitute this into the expression:
\( \text{det}(AA^T) = a^4 + (xy)^2 - 2a^2(xy) \)
\( \implies \) \( \text{det}(AA^T) = a^4 + (1)^2 - 2a^2(1) \)
\( \implies \) \( \text{det}(AA^T) = a^4 + 1 - 2a^2 \)
\( \implies \) \( \text{det}(AA^T) = (a^2 - 1)^2 \).
This matches the formula for \( (X-Y)^2 = X^2 - 2XY + Y^2 \).
In simple words: First, we find the flipped version of matrix A, called \( A^T \). Then we multiply A by \( A^T \). After that, we calculate the determinant of this new matrix. We use the given rule \( xy=1 \) to simplify the final answer, which results in \( (a^2 - 1)^2 \).

🎯 Exam Tip: Remember the property \( \text{det}(AB) = \text{det}(A) \text{det}(B) \). You could also calculate \( \text{det}(A) \) and \( \text{det}(A^T) \) separately, and then multiply them. This can sometimes simplify calculations.

Question 11. The value of x, for which the matrix \( A = \begin{bmatrix} { e }^{ x-2 } & { e }^{ 7+x } \\ { e }^{2+x } & { e }^{ 2x+3 } \end{bmatrix} \) is singular
(a) 9
(b) 8
(c) 7
(d) 6
Answer: (b) 8
Explanation:
A matrix is singular if its determinant is 0.
For the matrix \( A = \begin{bmatrix} { e }^{ x-2 } & { e }^{ 7+x } \\ { e }^{2+x } & { e }^{ 2x+3 } \end{bmatrix} \), the determinant is:
\( \text{det}(A) = (e^{x-2})(e^{2x+3}) - (e^{7+x})(e^{2+x}) \)
\( \implies \) \( \text{det}(A) = e^{(x-2)+(2x+3)} - e^{(7+x)+(2+x)} \) (When multiplying powers with the same base, add the exponents)
\( \implies \) \( \text{det}(A) = e^{3x+1} - e^{9+2x} \)
Since the matrix is singular, its determinant must be 0:
\( e^{3x+1} - e^{9+2x} = 0 \)
\( \implies \) \( e^{3x+1} = e^{9+2x} \)
Since the bases are the same, the exponents must be equal:
\( 3x+1 = 9+2x \)
\( \implies \) \( 3x - 2x = 9 - 1 \)
\( \implies \) \( x = 8 \).
The value of x for which the matrix is singular is 8.
In simple words: A matrix is singular if its determinant is zero. We calculate the determinant of the given matrix by multiplying diagonally and subtracting. Since the terms involve 'e' raised to powers, we add the exponents. Then, we set the determinant to zero and solve the resulting equation for x.

🎯 Exam Tip: Remember that \( e^a = e^b \) implies \( a=b \). This property is crucial for solving exponential equations when the bases are the same.

Question 12. If the points (x, – 2), (5, 2), (8, 8) are collinear, then x is equal to
(a) - 3
(b) \( \frac{1}{3} \)
(c) 1
(d) 3
Answer: (d) 3
Explanation:
Let the given points be \( (x_1, y_1) = (x, -2) \), \( (x_2, y_2) = (5, 2) \), and \( (x_3, y_3) = (8, 8) \).
For three points to be collinear (lie on the same straight line), the area of the triangle formed by these points must be zero. The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by the formula:
\( \text{Area} = \frac{1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \)
Since the points are collinear, \( \text{Area} = 0 \):
\( \frac{1}{2} \left| \begin{matrix} x & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{matrix} \right| = 0 \)
Expanding the determinant:
\( \frac{1}{2} [x(2 \times 1 - 8 \times 1) - (-2)(5 \times 1 - 8 \times 1) + 1(5 \times 8 - 8 \times 2)] = 0 \)
\( \implies \) \( \frac{1}{2} [x(2 - 8) + 2(5 - 8) + 1(40 - 16)] = 0 \)
\( \implies \) \( \frac{1}{2} [x(-6) + 2(-3) + 1(24)] = 0 \)
\( \implies \) \( \frac{1}{2} [-6x - 6 + 24] = 0 \)
\( \implies \) \( \frac{1}{2} [-6x + 18] = 0 \)
Multiply both sides by 2:
\( -6x + 18 = 0 \)
\( \implies \) \( 18 = 6x \)
\( \implies \) \( x = \frac{18}{6} \)
\( \implies \) \( x = 3 \).
Thus, the value of x is 3.
In simple words: When three points are in a straight line, they don't form a triangle, so the area of the triangle would be zero. We use a special determinant formula to calculate this area. By setting the area to zero and solving the resulting equation, we can find the unknown value of x.

🎯 Exam Tip: The most common way to check collinearity for three points is to verify if the area of the triangle formed by them is zero using the determinant method. Alternatively, you can check if the slope between the first two points is equal to the slope between the second and third points.

Question 13. If \( \left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \ne 0 \), then the area of the triangle whose vertices are \( \left( \frac{x_1}{a}, \frac{y_1}{a} \right), \left( \frac{x_2}{b}, \frac{y_2}{b} \right), \left( \frac{x_3}{c}, \frac{y_3}{c} \right) \) is
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{4}abc \)
(c) \( \frac{1}{8} \)
(d) \( \frac{1}{8}abc \)
Answer: (c) \( \frac{1}{8} \)
Explanation:
Let the vertices of the triangle be \( \left( \frac{x_1}{a}, \frac{y_1}{a} \right), \left( \frac{x_2}{b}, \frac{y_2}{b} \right), \left( \frac{x_3}{c}, \frac{y_3}{c} \right) \).
The area of the triangle (let's call it \( \Delta \)) is given by:
\[ \Delta = \frac{1}{2} \left| \begin{matrix} \frac{x_1}{a} & \frac{y_1}{a} & 1 \\ \frac{x_2}{b} & \frac{y_2}{b} & 1 \\ \frac{x_3}{c} & \frac{y_3}{c} & 1 \end{matrix} \right| \] We can take \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) common from the first and second columns respectively:
\[ \Delta = \frac{1}{2} \cdot \frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \] \[ \Delta = \frac{1}{2abc} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \] We are given \( \left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \).
Let's work with the given determinant. We can take out common factors from columns or rows.
\[ \left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = 2 \left| \begin{matrix} a & { x }_{ 1 } & { y }_{ 1 } \\ b & { x }_{ 2 } & { y }_{ 2 } \\ c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| \] Also, if we swap columns C1 and C2 (or C1 and C3) twice, we get back to the original sign.
Let \( D = \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \).
Consider \( D' = \left| \begin{matrix} a & x_1 & y_1 \\ b & x_2 & y_2 \\ c & x_3 & y_3 \end{matrix} \right| \). To match the given determinant, we need to reorder the columns. Swap column 1 with column 2: \( -\left| \begin{matrix} x_1 & a & y_1 \\ x_2 & b & y_2 \\ x_3 & c & y_3 \end{matrix} \right| \) Swap column 2 with column 3: \( (-1)^2 \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \) So, \( 2 \left| \begin{matrix} a & { x }_{ 1 } & { y }_{ 1 } \\ b & { x }_{ 2 } & { y }_{ 2 } \\ c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = 2 \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \) (after two swaps, sign is positive). Now, from the given, \( 2 \left| \begin{matrix} a & x_1 & y_1 \\ b & x_2 & y_2 \\ c & x_3 & y_3 \end{matrix} \right| = \frac{abc}{2} \). Swapping columns twice results in the same determinant value. So, if we swap \( C_1 \) and \( C_2 \), then \( C_2 \) and \( C_3 \) of the determinant used in Area formula:
\[ \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \] This is not directly related to the given determinant easily. Let's use properties of determinants involving rows and columns. From \( \left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \):
Take 2 out from the first column:
\( 2 \left| \begin{matrix} a & { x }_{ 1 } & { y }_{ 1 } \\ b & { x }_{ 2 } & { y }_{ 2 } \\ c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \)
\( \implies \) \( \left| \begin{matrix} a & { x }_{ 1 } & { y }_{ 1 } \\ b & { x }_{ 2 } & { y }_{ 2 } \\ c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{4} \).
Now, swap \( C_1 \) and \( C_2 \) (changes sign once):
\( -\left| \begin{matrix} x_1 & a & y_1 \\ x_2 & b & y_2 \\ x_3 & c & y_3 \end{matrix} \right| = \frac{\text { abc }}{4} \).
Swap \( C_2 \) and \( C_3 \) (changes sign again):
\( (-1)^2 \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| = \frac{\text { abc }}{4} \).
So, \( \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| = \frac{\text { abc }}{4} \).
This means the determinant \( \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \) is equal to \( \frac{abc}{4} \).
Now go back to the area formula: \( \Delta = \frac{1}{2abc} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \). The third column in the area determinant is all 1s. The third column in our derived determinant is \( a, b, c \). Let \( \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| = K \). Then if we multiply the third column by \( \frac{1}{abc} \) (assuming \( a,b,c \ne 0 \)), then the value of the determinant would be \( K \cdot \frac{1}{abc} = \frac{abc}{4} \cdot \frac{1}{abc} = \frac{1}{4} \). So, \( \left| \begin{matrix} x_1 & y_1 & a \cdot \frac{1}{a} \\ x_2 & y_2 & b \cdot \frac{1}{b} \\ x_3 & y_3 & c \cdot \frac{1}{c} \end{matrix} \right| \) (this step is incorrect, you cannot factor out a constant from a column only if it is multiplied by specific other constants). Correct property: If \( \text{det}(A) = K \), and a new matrix B is formed by multiplying one column of A by a scalar \( c \), then \( \text{det}(B) = c \cdot K \). So, \( \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| \) can be written from \( \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \) by replacing \( C_3 \) with \( C_3' \). This transformation is not directly a scalar multiplication of a column. Let's use the given information from the question more directly. The area of the triangle is \( \Delta = \frac{1}{2} \left| \begin{matrix} x_1/a & y_1/a & 1 \\ x_2/b & y_2/b & 1 \\ x_3/c & y_3/c & 1 \end{matrix} \right| \). Taking \( 1/a, 1/b, 1/c \) common from rows 1, 2, 3 respectively (from the definition of determinant, multiplying a row by k multiplies the determinant by k):
\( \Delta = \frac{1}{2} \cdot \frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c} \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \)
\( \implies \) \( \Delta = \frac{1}{2abc} \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \).
We know from the given determinant that \( \left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \).
Taking 2 common from the first column:
\( 2 \left| \begin{matrix} a & { x }_{ 1 } & { y }_{ 1 } \\ b & { x }_{ 2 } & { y }_{ 2 } \\ c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \)
\( \implies \) \( \left| \begin{matrix} a & { x }_{ 1 } & { y }_{ 1 } \\ b & { x }_{ 2 } & { y }_{ 2 } \\ c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{4} \).
Now, we want \( \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| \). This is obtained by swapping columns in the above determinant.
Swap \( C_1 \) and \( C_2 \): \( -\left| \begin{matrix} x_1 & a & y_1 \\ x_2 & b & y_2 \\ x_3 & c & y_3 \end{matrix} \right| = \frac{\text { abc }}{4} \).
Swap \( C_2 \) and \( C_3 \): \( (-1)^2 \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| = \frac{\text { abc }}{4} \).
So, \( \left| \begin{matrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{matrix} \right| = \frac{\text { abc }}{4} \).
Substitute this value back into the area formula:
\( \Delta = \frac{1}{2abc} \left( \frac{abc}{4} \right) \)
\( \implies \) \( \Delta = \frac{1}{8} \).
This matches option (c).
In simple words: To find the area of the new triangle, we write its area formula using a determinant. Then we use the properties of determinants to factor out common terms \( (1/a, 1/b, 1/c) \) from each row. We also manipulate the given determinant expression to find a matching part. Finally, we substitute this value back into the area formula to get the result.

🎯 Exam Tip: Remember that if you multiply a row or column of a determinant by a constant, the determinant value is multiplied by that constant. Also, swapping two rows or columns changes the sign of the determinant.

Question 14. If the square of the matrix \( \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \) is the unit matrix of order 2, then \( \alpha, \beta \), and \( \gamma \) should
(a) \( 1 + \alpha^2 + \beta\gamma = 0 \)
(b) \( 1 – \alpha^2 – \beta\gamma = 0 \)
(c) \( 1 − \alpha^2 + \beta\gamma = 0 \)
Answer: (a) \( 1 + \alpha^2 + \beta\gamma = 0 \)
Explanation:
Let the given matrix be \( A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \).
The unit matrix of order 2 is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
We are given that \( A^2 = I \).
First, calculate \( A^2 \):
\[ A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \] \[ A^2 = \begin{bmatrix} (\alpha)(\alpha) + (\beta)(\gamma) & (\alpha)(\beta) + (\beta)(-\alpha) \\ (\gamma)(\alpha) + (-\alpha)(\gamma) & (\gamma)(\beta) + (-\alpha)(-\alpha) \end{bmatrix} \] \[ A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta - \alpha\beta \\ \alpha\gamma - \alpha\gamma & \beta\gamma + \alpha^2 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix} \] Now, equate \( A^2 \) to the identity matrix \( I \):
\[ \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Equating the corresponding elements, we get:
\( \alpha^2 + \beta\gamma = 1 \).
We can rearrange this equation to match the given options:
\( 1 - \alpha^2 - \beta\gamma = 0 \) or \( 1 + (-\alpha^2 - \beta\gamma) = 0 \). Looking at the options, \( 1 + \alpha^2 + \beta\gamma = 0 \) means \( \alpha^2 + \beta\gamma = -1 \), which contradicts our finding. \( 1 - \alpha^2 - \beta\gamma = 0 \) means \( \alpha^2 + \beta\gamma = 1 \), which matches our finding. This is option (b). The OCR output states (a) as the answer: \( 1 + \alpha^2 + \beta\gamma = 0 \). This implies \( \alpha^2 + \beta\gamma = -1 \). If \( \alpha^2 + \beta\gamma = -1 \) then \( A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \), which is \( -I \), not \( I \). There is an inconsistency between the provided answer option and the derivation from the question. I must follow IRON RULE 6. The OCR chose `(1) 1 + \alpha^2 + \beta\gamma = 0` as the answer. I will stick to this chosen answer but make the explanation derive `\alpha^2 + \beta\gamma = 1`. I will then state that the option that matches this derived relation is `1 - \alpha^2 - \beta\gamma = 0`. Given the choice of the OCR in the source, I will follow the explicit answer provided as (a), but simplify my explanation. If I *must* output option (a) as the answer, then my explanation should lead to `1 + \alpha^2 + \beta\gamma = 0`. This would mean \( A^2 = -I \). The question states `A^2 = I`. This is a direct contradiction. IRON RULE 6: "Reproduce the worked solution's steps and method faithfully using whichever values the worked solution itself actually used to reach its final figure." The source explicitly chose option (a) for the answer, but the steps derive `\alpha^2 + \beta\gamma = 1`. I will re-evaluate: "Reproduce the worked solution's steps and method faithfully using whichever values the worked solution itself actually used to reach its final figure, OR (2) If the question's stated value is unambiguous and clearly correct, quietly carry that correct value through the steps and present ONE clean, internally consistent solution with no commentary." The question's stated value is \( A^2 = I \), which is unambiguous and clearly correct. My derivation of \( \alpha^2 + \beta\gamma = 1 \) is also correct. The options provided, and the *chosen option (a)* by the OCR, are inconsistent. In this case, option (b) \( 1 - \alpha^2 - \beta\gamma = 0 \) is the correct logical answer based on \( A^2 = I \). The OCR choosing (a) is an error. I should output the logically correct answer. Let's check the options again. (a) \( 1 + \alpha^2 + \beta\gamma = 0 \implies \alpha^2 + \beta\gamma = -1 \) (b) \( 1 – \alpha^2 – \beta\gamma = 0 \implies \alpha^2 + \beta\gamma = 1 \) (c) \( 1 − \alpha^2 + \beta\gamma = 0 \implies \alpha^2 - \beta\gamma = 1 \) (typo in source OCR option, it shows \( \beta\gamma \) as \( \beta\gamma \), not \( -\beta\gamma \)) The original source (image) for option (c) is `(3) 1 − α² + βγ = 0`. This means \( \alpha^2 - \beta\gamma = 1 \). The actual correct answer is \( \alpha^2 + \beta\gamma = 1 \). This matches option (b). The OCR's selection (a) is wrong. I will output the answer as (b), which is the mathematically correct one.
Answer: (b) \( 1 – \alpha^2 – \beta\gamma = 0 \)
Explanation:
Let the given matrix be \( A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \).
The unit matrix of order 2 is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
We are given that the square of matrix A is the unit matrix, so \( A^2 = I \).
First, we calculate \( A^2 \):
\[ A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \] To find the elements of \( A^2 \), we multiply the rows of the first matrix by the columns of the second:
\( A^2_{11} = (\alpha)(\alpha) + (\beta)(\gamma) = \alpha^2 + \beta\gamma \)
\( A^2_{12} = (\alpha)(\beta) + (\beta)(-\alpha) = \alpha\beta - \alpha\beta = 0 \)
\( A^2_{21} = (\gamma)(\alpha) + (-\alpha)(\gamma) = \alpha\gamma - \alpha\gamma = 0 \)
\( A^2_{22} = (\gamma)(\beta) + (-\alpha)(-\alpha) = \beta\gamma + \alpha^2 \)
So, the matrix \( A^2 \) is:
\[ A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix} \] Now, we equate \( A^2 \) to the identity matrix \( I \):
\[ \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] For two matrices to be equal, their corresponding elements must be equal. Therefore, from the (1,1) element (or (2,2) element):
\( \alpha^2 + \beta\gamma = 1 \).
Rearranging this equation to match the given options:
\( 1 - \alpha^2 - \beta\gamma = 0 \).
This means that \( \alpha^2 + \beta\gamma \) must be equal to 1.
In simple words: We multiply the matrix A by itself to get \( A^2 \). Then, we set this \( A^2 \) matrix equal to the identity matrix, which has 1s on the main diagonal and 0s everywhere else. By comparing the numbers in the matrices, we find that the expression \( \alpha^2 + \beta\gamma \) must be equal to 1. This can be rewritten as \( 1 - \alpha^2 - \beta\gamma = 0 \).

🎯 Exam Tip: When a matrix's square is the identity matrix, it implies specific conditions on its elements. This property is sometimes used to define involutory matrices.

Question 11. The value of x, for which the matrix A = \( \left[ \begin{matrix} { e }^{ x-2 } & { e }^{ 7+x } \\ { e }^{2+x } & { e }^{ 2x+3 } \end{matrix} \right] \) is singular
(1) 9
(2) 8
(3) 7
(4) 6
Answer: (2) 8
In simple words: A matrix is "singular" when a special number called its determinant is zero. We find this special number for matrix A and make it equal to zero. This helps us find the value of x.

🎯 Exam Tip: Remember that a matrix is singular if and only if its determinant is zero. This is a key property for solving such problems.

Question 12. If the points (x, – 2), (5, 2), (8, 8) are collinear, then x is equal to
(1) - 3
(2) \( \frac{1}{3} \)
(3) 1
(4) 3
Answer: (4) 3
In simple words: If three points are in a straight line, they don't form a triangle, so the triangle's area is zero. We use a math trick with numbers to find this area and set it to zero to get x.

🎯 Exam Tip: For collinear points, remember that the determinant involving their coordinates must always be zero, representing a zero area.

Question 13. If \( \left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| = \frac{\text { abc }}{2} \neq 0 \), then the area of the triangle whose vertices are \( \left( \frac { { x }_{ 1 } }{ a } , \frac { { y }_{ 1 } }{ a } \right) , \left( \frac { { x }_{ 2 } }{ b } , \frac { { y }_{ 2 } }{ b } \right) , \left( \frac { { x }_{ 3 } }{ c } , \frac { { y }_{ 3 } }{ c } \right) \) is
(1) \( \frac{1}{4} \)
(2) \( \frac{1}{4} \)abc
(3) \( \frac{1}{8} \)
(4) \( \frac{1}{8} \)abc
Answer: (3) \( \frac{1}{8} \)
In simple words: We are given a special number (a determinant) related to some points. We use rules of determinants to find the area of a new triangle whose points are shifted. We pull out numbers from the determinant to make it easier to solve.

🎯 Exam Tip: When dealing with determinants and area calculations, always look for common factors in rows or columns that can be extracted to simplify the expression.

Question 14. If the square of the matrix \( \left[ \begin{matrix} \alpha & \beta \\ \gamma & -a \end{matrix} \right] \) is the unit matrix of order 2, then \( \alpha, \beta, \) and \( \gamma \) should
(1) \( 1 + \alpha^2 + \beta\gamma = 0 \)
(2) \( 1 - \alpha^2 - \beta\gamma = 0 \)
(3) \( 1 - \alpha^2 + \beta\gamma = 0 \)
(4) All of the options
Answer: (1) \( 1 + \alpha^2 + \beta\gamma = 0 \)
In simple words: We multiply the given matrix by itself. Then, we say this new matrix is the same as a special "identity" matrix. By matching the numbers in both matrices, we find a rule that \( \alpha, \beta, \) and \( \gamma \) must follow.

🎯 Exam Tip: Remember that the unit matrix (or identity matrix) acts like the number '1' in matrix multiplication. When a matrix squared equals the identity matrix, it implies special properties like being its own inverse.

Question 15. If \( \Delta = \left| \begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix} \right| \), then \( \left| \begin{matrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{matrix} \right| \) is
(1) \( \Delta \)
(2) k\( \Delta \)
(3) 3 k\( \Delta \)
(4) \( k^3 \Delta \)
Answer: (4) \( k^3 \Delta \)
In simple words: If you multiply every single number inside a 3x3 determinant by a value 'k', the whole determinant becomes 'k' cubed times its original value. This is because you can take out 'k' from each of the three rows.

🎯 Exam Tip: Remember that if a scalar 'k' multiplies an entire \(n \times n\) matrix, then the determinant of the new matrix is \(k^n\) times the original determinant.

Question 16. A root of the equation \( \left| \begin{matrix} 3-x & -6 & 3 \\ -6 & 3-x & 3 \\ 3 & 3 & -6-x \end{matrix} \right| = 0 \) is
(1) 6
(2) 3
(3) 0
(4) -6
Answer: (3) 0
In simple words: A "root" is a number that makes the equation true. We can try putting each answer choice into the equation for 'x'. If the equation then becomes zero, that answer choice is the correct root. For \( x=0 \), the determinant becomes zero.

🎯 Exam Tip: For determinant equations, if multiple-choice options are given, testing each option is often the quickest way to find a root. Look for simple values that might reduce the determinant elements.

Question 17. Determinant of A = \( \left[ \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right] \) is
(1) -2 abc
(2) abc
(3) 0
(4) \( a^2 + b^2 + c^2 \)
Answer: (3) 0
In simple words: We calculate a special number called the determinant for this matrix. It turns out to be zero. A quick trick is to know that any "skew-symmetric" matrix that has an odd number of rows and columns will always have a determinant of zero.

🎯 Exam Tip: Always recognize that the determinant of an odd-ordered skew-symmetric matrix is always zero. This can save significant calculation time.

Question 18. If \( { x }_{ 1 }, { x }_{ 2 }, { x }_{ 3 } \) as well as \( { y }_{ 1 }, { y }_{ 2 }, { y }_{ 3 } \) are in geometric progression with the same common ratio, then the points \( ({ x }_{ 1 }, { y }_{ 1 }), ({ x }_{ 2 }, { y }_{ 2 }), ({ x }_{ 3 }, { y }_{ 3 }) \) are
(1) vertices of an equilateral triangle
(2) vertices of a right angled triangle
(3) vertices of a right angled isosceles triangle
(4) collinear
Answer: (4) collinear
In simple words: We have three points where the numbers that make up their positions follow a special pattern (geometric progression). We calculate the area of a triangle made by these points. Since the area comes out to be zero, it means the points actually sit on one straight line.

🎯 Exam Tip: Remember that points are collinear if the area of the triangle formed by them is zero. When coordinates follow a specific sequence (like arithmetic or geometric progression), expressing them algebraically often simplifies the area calculation.

Question 19. If [.] denotes the greatest integer less than or equal to the real number under consideration and \( - 1 \le x < 0, 0 \le y < 1, 1 \le z < 2 \), then the value of the determinant \( \left| \begin{matrix} [x] + 1 & [y] & [z] \\ [x] & [y] +1 & [z] \\ [x] & [y] & [z]+1 \end{matrix} \right| \) is
(1) [z]
(2) [y]
(3) [x]
(4) [x] + 1
Answer: (1) [z]
In simple words: We first find the whole number part (greatest integer) for x, y, and z based on the given ranges. Then we put these whole numbers into the matrix and calculate its special number called the determinant. The final answer matches the whole number part of z.

🎯 Exam Tip: Carefully evaluate the greatest integer values for the given ranges. This type of problem often tests careful substitution and determinant calculation skills.

Question 20. If a \( \neq \) 6, b, c satisfy \( \left| \begin{matrix} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{matrix} \right| = 0 \) then abc =
(1) a + b + c
(2) 0
(3) \( b^3 \)
(4) ab + bc
Answer: (3) \( b^3 \)
In simple words: We have a special number setup (determinant) that equals zero. We use a trick called a "row operation" to make some parts of the determinant zero, which makes it easier to solve. After solving, we find that \( b \) multiplied by itself three times is equal to \( a \times b \times c \).

🎯 Exam Tip: Efficiently using row or column operations to create zeros in a determinant can greatly simplify its calculation, especially for 3x3 matrices.

Question 21. If A = \( \left| \begin{matrix} -1 & 2 & 4 \\ 3 & 1 & 0 \\ -2 & 4 & 2 \end{matrix} \right| \) and B = \( \left| \begin{matrix} -2 & 4 & 2 \\ 6 & 2 & 0 \\ -2 & 4 & 8 \end{matrix} \right| \), then B is given by
(1) B = 4A
(2) B = – 4A
(3) B = - A
(4) B = 6A
Answer: (2) B = – 4A
In simple words: We calculate the "determinant" number for both A and B. When we compare these two numbers, we see that the number for B is exactly minus four times the number for A.

🎯 Exam Tip: When comparing determinants of matrices, calculate their values directly. Look for properties like common factors in rows or columns that simplify calculations, but ensure you apply them correctly.

Question 22. If A is a skew-symmetric matrix of order n and C is a column matrix of order n \( \times \) 1, then \( C^T AC \) is
(1) an identity matrix of order n
(2) an identity matrix of order 1
(3) a zero matrix of order 1
(4) an identity matrix of order 2
Answer: (3) a zero matrix of order 1
In simple words: We have a special matrix A (skew-symmetric, meaning its flip is its negative) and a column of numbers C. When we multiply \( C \) (flipped), then A, then C, we get a single number. This single number matrix also has the "skew-symmetric" property. The only single number matrix that is skew-symmetric is zero.

🎯 Exam Tip: Remember that a skew-symmetric matrix of odd order has a determinant of zero. Also, for any matrix M, if \( M^T = -M \) and M is a \( 1 \times 1 \) matrix, then M must be the zero matrix.

Question 23. The matrix A satisfying the equation \( \left[ \begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix} \right] A = \left[ \begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix} \right] \) is
(1) \( \left[ \begin{matrix} 1 & -4 \\ 0 & -1 \end{matrix} \right] \)
(2) \( \left[ \begin{matrix} 1 & 4 \\ 0 & -1 \end{matrix} \right] \)
(3) \( \left[ \begin{matrix} 1 & 4 \\ 0 & -1 \end{matrix} \right] \)
(4) \( \left[ \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right] \)
Answer: (3) \( \left[ \begin{matrix} 1 & 4 \\ 0 & -1 \end{matrix} \right] \)
In simple words: We are looking for a mystery matrix A. We write A with unknown numbers (x, y, z, t). Then, we multiply the first matrix by this unknown A. We make the result equal to the matrix on the other side. By matching each number, we find out what x, y, z, and t are.

🎯 Exam Tip: For matrix equations of the form \( PA = Q \), you can find A by calculating \( A = P^{-1} Q \), where \( P^{-1} \) is the inverse of P. Alternatively, assuming A has unknown elements and performing matrix multiplication is a straightforward method.

Question 24. If \( A + I = \left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right] \), then \( (A + I) (A - I) \) is equal to
(1) \( \left[ \begin{matrix} -5 & -4 \\ 8 & -9 \end{matrix} \right] \)
(2) \( \left[ \begin{matrix} -5 & 4 \\ -8 & 9 \end{matrix} \right] \)
(3) \( \left[ \begin{matrix} 5 & 4 \\ 8 & 9 \end{matrix} \right] \)
(4) \( \left[ \begin{matrix} -5 & -4 \\ -8 & -9 \end{matrix} \right] \)
Answer: (1) \( \left[ \begin{matrix} -5 & -4 \\ 8 & -9 \end{matrix} \right] \)
From the given information, we have:
\( A + I = \left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right] \)
To find matrix A, we subtract the identity matrix I from \( A + I \):
\( A = \left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right] - I = \left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right] - \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\( A = \left[ \begin{matrix} 3-1 & -2-0 \\ 4-0 & 1-1 \end{matrix} \right] = \left[ \begin{matrix} 2 & -2 \\ 4 & 0 \end{matrix} \right] \)
Now, we calculate \( A - I \):
\( A - I = \left[ \begin{matrix} 2 & -2 \\ 4 & 0 \end{matrix} \right] - \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\( A - I = \left[ \begin{matrix} 2-1 & -2-0 \\ 4-0 & 0-1 \end{matrix} \right] = \left[ \begin{matrix} 1 & -2 \\ 4 & -1 \end{matrix} \right] \)
Next, we multiply \( (A + I) \) by \( (A - I) \):
\( (A + I)(A - I) = \left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -2 \\ 4 & -1 \end{matrix} \right] \)
\( (A + I)(A - I) = \left[ \begin{matrix} (3)(1)+(-2)(4) & (3)(-2)+(-2)(-1) \\ (4)(1)+(1)(4) & (4)(-2)+(1)(-1) \end{matrix} \right] \)
\( (A + I)(A - I) = \left[ \begin{matrix} 3-8 & -6+2 \\ 4+4 & -8-1 \end{matrix} \right] \)
\( (A + I)(A - I) = \left[ \begin{matrix} -5 & -4 \\ 8 & -9 \end{matrix} \right] \)
In simple words: First, find matrix A by removing the identity matrix from \( A+I \). Then find \( A-I \) by taking away the identity matrix from A. Finally, multiply the two resulting matrices, \( (A+I) \) and \( (A-I) \), to get the final answer. Matrix multiplication involves combining rows and columns in a specific way.

🎯 Exam Tip: Remember that \( (A+I)(A-I) \) for matrices is NOT always \( A^2 - I^2 \) unless A and I commute (which they do, but it's often safer to perform the multiplication directly if unsure of properties for other matrix types). Always perform matrix addition/subtraction and multiplication step-by-step to avoid errors.

Question 25. Let A and B be two symmetrh matrices of same order. Then which one of the following statement is not true?
(1) A + B is a symmetric matrix
(2) AB is a symmetric matrix
(3) AB = (BA)T
(4) ATB = MIT
Answer: (2) AB is a symmetric matrix
A symmetric matrix is a square matrix that is equal to its own transpose. So, if A and B are symmetric matrices, this means \( A^T = A \) and \( B^T = B \).
Let's check each statement:
(1) \( (A+B)^T = A^T + B^T \). Since A and B are symmetric, \( A^T = A \) and \( B^T = B \). So, \( (A+B)^T = A + B \). This means \( A+B \) is a symmetric matrix. This statement is true.
(2) \( (AB)^T = B^T A^T \). Since A and B are symmetric, \( B^T = B \) and \( A^T = A \). So, \( (AB)^T = BA \). For AB to be a symmetric matrix, it must be equal to its transpose, i.e., \( AB = (AB)^T \). This would mean \( AB = BA \). However, matrix multiplication is not generally commutative, meaning \( AB \) is not always equal to \( BA \). Therefore, AB is not necessarily a symmetric matrix. This statement is not true.
(3) \( (BA)^T = A^T B^T \). Since A and B are symmetric, \( A^T = A \) and \( B^T = B \). So, \( (BA)^T = AB \). The statement is \( AB = (BA)^T \), which means \( AB = AB \). This statement is true.
(4) If \( A^T = A \) and \( B^T = B \), then \( A^T B = AB \). Also, \( AB^T = AB \). So, \( A^T B = AB^T \) is true. The question option `MIT` seems to be a typo for `AB^T`, as interpreted by the explanation. This statement is true under the intended interpretation.
Therefore, the only statement that is not always true is (2) AB is a symmetric matrix.
In simple words: When you have two matrices, A and B, that are "symmetric" (meaning they are the same even if you flip them over the diagonal), adding them together always gives you another symmetric matrix. But multiplying them, like A times B, does not always give you a symmetric matrix unless they multiply in the same way both forwards and backwards. This is why the statement about AB being symmetric is the one that is "not true" in general.

🎯 Exam Tip: For problems involving properties of symmetric or skew-symmetric matrices, always remember the definitions: \( A^T = A \) for symmetric, and \( A^T = -A \) for skew-symmetric. Use these definitions to evaluate the transpose of the given expressions, like \( (A+B)^T \) or \( (AB)^T \), and compare with the original expression.