CBSE Class 12 Mathematics Matrices and Determinants Assignment Set 10

Question. Find \(x\) and \(y\) if: \(2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
Answer: Multiply the first matrix by 2:
\(\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
Add the two matrices on the left-hand side:
\(\begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
Equating corresponding elements:
\(2 + y = 5 \implies y = 3\)
\(2x + 2 = 8 \implies 2x = 6 \implies x = 3\)
Thus, \(x = 3\) and \(y = 3\).
In simple words: Multiply, add, and compare corresponding matrix entries to find the variables.
Exam Tip: Always perform scalar multiplication before adding matrices.

Question. Evaluate: \(\begin{vmatrix} a + ib & c + id \\ -c + id & a - ib \end{vmatrix} \)
Answer: Calculate the determinant by cross-multiplication:
\(= (a + ib)(a - ib) - (c + id)(-c + id)\)
\(= (a^2 - (ib)^2) - ((id + c)(id - c))\)
\(= (a^2 + b^2) - ((id)^2 - c^2)\)
\(= (a^2 + b^2) - (-d^2 - c^2)\)
\(= a^2 + b^2 + c^2 + d^2\).
In simple words: Use the property \((x+y)(x-y) = x^2 - y^2\) and \(i^2 = -1\) to simplify the product.
Exam Tip: Remember that \(i^2 = -1\) transforms negative signs into positive terms.

Question. Find the cofactor of diagonal elements in the following: \(\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \), also find \(a_{23} \cdot A_{23}\).
Answer: The diagonal elements are \(a_{11} = 2\), \(a_{22} = 0\), and \(a_{33} = -7\).
Cofactor formula is \(A_{ij} = (-1)^{i+j} M_{ij}\):
Cofactor of \(a_{11}\) (i.e., \(A_{11}\)) = \((-1)^{1+1} \begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = 1 \cdot (0(-7) - 5(4)) = -20\).
Cofactor of \(a_{22}\) (i.e., \(A_{22}\)) = \((-1)^{2+2} \begin{vmatrix} 2 & 5 \\ 1 & -7 \end{vmatrix} = 1 \cdot (2(-7) - 1(5)) = -19\).
Cofactor of \(a_{33}\) (i.e., \(A_{33}\)) = \((-1)^{3+3} \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = 1 \cdot (2(0) - 6(-3)) = 18\).
For \(a_{23} \cdot A_{23}\):
Element \(a_{23} = 4\).
Cofactor \(A_{23} = (-1)^{2+3} \begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = -1 \cdot (2(5) - 1(-3)) = -(10 + 3) = -13\).
Therefore, \(a_{23} \cdot A_{23} = 4 \times (-13) = -52\).
In simple words: Delete the row and column of each target element to find the minor, apply the sign rule, and multiply.
Exam Tip: Be very careful with the alternating signs \((-1)^{i+j}\) when calculating cofactors.

Question. For what value of \(x\), is the following matrix singular? \(\begin{bmatrix} 3 - 2x & x + 1 \\ 2 & 4 \end{bmatrix} \)
Answer: A matrix is singular if its determinant is equal to 0.
\(\begin{vmatrix} 3 - 2x & x + 1 \\ 2 & 4 \end{vmatrix} = 0 \)
\(4(3 - 2x) - 2(x + 1) = 0\)
\(12 - 8x - 2x - 2 = 0\)
\(10 - 10x = 0\)
\(10x = 10 \implies x = 1\).
In simple words: Set the determinant of the matrix to zero and solve for \(x\).
Exam Tip: "Singular matrix" always means determinant equals zero (\(|A| = 0\)).

Question. A matrix \(A\), of order \(3 \times 3\), has determinant 4. Find the value of \(|3A|\).
Answer: Using the property for a square matrix \(A\) of order \(n\): \(|kA| = k^n |A|\).
Here, \(k = 3\), \(n = 3\), and \(|A| = 4\).
\(|3A| = 3^3 \cdot |A|\)
\(|3A| = 27 \times 4 = 108\).
In simple words: Pulling a scalar \(k\) out of an \(n \times n\) determinant raises it to the power of \(n\).
Exam Tip: Do not simply multiply the determinant by 3; remember to raise 3 to the power of the matrix order.

Question. Construct a \(2 \times 3\) matrix \(A = [a_{ij}]\) whose elements are given by \(a_{ij} = \frac{3 + i}{2 - j}\).
Answer: A general \(2 \times 3\) matrix is given by:
\(A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \)
Calculate each element using \(a_{ij} = \frac{3 + i}{2 - j}\):
\(a_{11} = \frac{3 + 1}{2 - 1} = \frac{4}{1} = 4\)
\(a_{12} = \frac{3 + 1}{2 - 2} = \frac{4}{0}\) (Undefined; assuming standard typo in school worksheets or restricted domains, but standard procedure follows formula substitution)
Let us write standard values assuming valid indices or evaluate formally as per formula: Since denominator becomes 0 at \(j = 2\), let us check if the question implies standard forms. Proceeding with standard formula output:
\(a_{13} = \frac{3 + 1}{2 - 3} = \frac{4}{-1} = -4\)
\(a_{21} = \frac{3 + 2}{2 - 1} = \frac{5}{1} = 5\)
\(a_{22} = \frac{3 + 2}{2 - 2} = \text{Undefined}\)
\(a_{23} = \frac{3 + 2}{2 - 3} = \frac{5}{-1} = -5\)
In simple words: Substitute the row index \(i\) and column index \(j\) into the given formula for each position.
Exam Tip: Double check indices \(i\) (row) and \(j\) (column) before calculating.

Question. If \(\begin{vmatrix} x & x \\ 1 & x \end{vmatrix} = \begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix} \), find the value of \(x\).
Answer: Evaluate both determinants:
Left-hand side: \(x(x) - (1)(x) = x^2 - x\)
Right-hand side: \(3(2) - 1(4) = 6 - 4 = 2\)
Equating both sides:
\(x^2 - x = 2\)
\(x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0\)
\(x = 2\) or \(x = -1\).
In simple words: Expand both determinants separately and solve the resulting quadratic equation.
Exam Tip: Do not cancel \(x\) across matrices; always evaluate determinants to numbers first.

Question. If matrix \(A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\), find \(A A^T\).
Answer: Matrix \(A\) is of order \(1 \times 3\).
The transpose matrix \(A^T\) is of order \(3 \times 1\):
\(A^T = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
Now find the product \(A A^T\):
\(A A^T = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
\(= [1(1) + 2(2) + 3(3)]\)
\(= [1 + 4 + 9] = [14]\).
In simple words: Multiply a \(1 \times 3\) row vector by its corresponding \(3 \times 1\) column vector to get a \(1 \times 1\) matrix.
Exam Tip: The product of a matrix and its transpose results in a 1x1 matrix here, write it inside brackets or as a scalar depending on context.

Question. Write the value of the determinant \(\begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{vmatrix} \).
Answer: Factor out \(3x\) from the third row:
\(= 3x \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4 \end{vmatrix} \)
Since Row 1 and Row 3 are identical (\(R_1 = R_3\)), the value of the determinant is 0.
\(= 3x \times 0 = 0\).
In simple words: If any two rows or columns of a determinant are identical, its value is zero.
Exam Tip: Always look for proportional rows or columns before expanding large determinants.

Question. If \(A\) is an invertible matrix of order 3 and \(|A| = 5\), then find \(|\text{adj} A|\)
Answer: Using the standard property for the determinant of the adjoint: \(|\text{adj} A| = |A|^{n-1}\).
Here, order \(n = 3\) and \(|A| = 5\).
\(|\text{adj} A| = 5^{3-1} = 5^2 = 25\).
In simple words: The determinant of the adjoint of a matrix equals the determinant of the matrix raised to the power of \(n-1\).
Exam Tip: Memorize the direct formula \(|\text{adj} A| = |A|^{n-1}\) for quick marks.

Question. Find \(x\) for which \(\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = 0 \)
Answer: Expand the determinant:
\((x)(2x) - (4)(2) = 0\)
\(2x^2 - 8 = 0\)
\(2x^2 = 8\)
\(x^2 = 4\)
\(x = \pm 2\).
In simple words: Cross-multiply and solve the simple quadratic equation for \(x\).
Exam Tip: Do not forget both positive and negative roots when taking square roots (\(x = \pm 2\)).

Question. Write the value of the determinant: \(\begin{vmatrix} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{vmatrix} \)
Answer: Apply the row operation \(R_1 \rightarrow R_1 + R_2 + R_3\):
Sum of elements in the first row:
\((a - b + b - c + c - a) = 0\)
Similarly for all columns when added, or noting that adding all rows yields zero in every element of the first row:
\(R_1 = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}\)
Since all elements of the first row are zero, the value of the determinant is 0.
In simple words: Adding all rows together gives a row of zeros, making the entire determinant zero.
Exam Tip: Cyclic determinants where row sums equal zero always evaluate to 0.

Question. Evaluate: \(\begin{vmatrix} 7 & -2 \\ -10 & 5 \end{vmatrix} \) (Note: leading factor 2 or standard determinant format)
Answer: Evaluating the determinant component:
\(= (7)(5) - (-2)(-10)\)
\(= 35 - (20) = 15\).
In simple words: Multiply main diagonal elements and subtract the product of off-diagonal elements.
Exam Tip: Watch out for double negatives when multiplying negative numbers.

Question. If \(A = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \), find \(A + A'\).
Answer: Find the transpose matrix \(A'\):
\(A' = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} \)
Add \(A\) and \(A'\):
\(A + A' = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} + \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} \)
\(= \begin{bmatrix} 3+3 & 4+2 \\ 2+4 & 3+3 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 6 & 6 \end{bmatrix} \).
In simple words: Add a matrix to its transpose by summing corresponding elements.
Exam Tip: The sum of a square matrix and its transpose is always a symmetric matrix.

Question. If \(A\) is a non-singular matrix of order 3 and \(|\text{adj} A| = |A|^K\), write the value of \(K\).
Answer: We know that for any non-singular square matrix of order \(n\), \(|\text{adj} A| = |A|^{n-1}\).
Given order \(n = 3\), so \(|\text{adj} A| = |A|^{3-1} = |A|^2\).
Comparing with \(|A|^K\), we get \(K = 2\).
In simple words: The exponent \(K\) in the adjoint determinant formula is always \(n-1\).
Exam Tip: For order \(n\), \(K = n - 1\).

Question. If \(A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \), then for what value of \(\alpha\) is \(A\) an identity matrix?
Answer: An identity matrix of order 2 is \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Equating corresponding elements of \(A\) to the identity matrix:
\(\cos\alpha = 1\) and \(\sin\alpha = 0\)
The common value of \(\alpha\) in \([0, 2\pi)\) satisfying both equations is \(\alpha = 0\) (or \(2n\pi\)).
In simple words: Find the angle where cosine is 1 and sine is 0.
Exam Tip: Check that the angle satisfies both trigonometric equations simultaneously.

Question. If \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \), find the value of \(k\).
Answer: Multiply the two matrices on the left-hand side:
\(\begin{bmatrix} (1)(3) + (2)(2) & (1)(1) + (2)(5) \\ (3)(3) + (4)(2) & (3)(1) + (4)(5) \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix}\)
Simplify the entries:
\(\begin{bmatrix} 3 + 4 & 1 + 10 \\ 9 + 8 & 3 + 20 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ 17 & 23 \end{bmatrix} \)
Comparing the element corresponding to \(k\) (second row, first column):
\(k = 17\).
In simple words: Perform matrix multiplication on the left side and match the resulting element to \(k\).
Exam Tip: You only need to calculate the specific row-column product that yields \(k\) to save time.

Question. Write the adjoint of \(A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \)
Answer: For a \(2 \times 2\) matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint is found by swapping the diagonal elements and changing the signs of the off-diagonal elements: \(\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
Here, \(a = 2, b = -1, c = 4, d = 3\):
\(\text{adj} A = \begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} \).
In simple words: Swap the main diagonal numbers and negate the other two numbers for a \(2 \times 2\) matrix.
Exam Tip: Save precious exam time by using the direct shortcut for \(2 \times 2\) adjoint matrices.

Question. A is a square matrix of order 3 and \(|A| = 7\). Write the value of \(|\text{adj} A|\) and \(|3A|\).
Answer: 1. For \(|\text{adj} A|\):
\(|\text{adj} A| = |A|^{n-1} = 7^{3-1} = 7^2 = 49\).
2. For \(|3A|\):
\(|3A| = 3^n |A| = 3^3 \times 7 = 27 \times 7 = 189\).
In simple words: Apply the standard properties \(|\text{adj} A| = |A|^{n-1}\) and \(|kA| = k^n |A|\).
Exam Tip: Keep separate formulas clear in mind: adjoint power is \(n-1\), scalar multiple power is \(n\).

Question. If A and B are matrices of order \(3 \times 4\) and \(4 \times 3\) respectively, find the order of matrix \((AB)\) and \((BA)\).
Answer: - For the product \(AB\): Order of \(A\) is \(3 \times 4\) and order of \(B\) is \(4 \times 3\). Since inner dimensions match (4 = 4), the order of \(AB\) is \(3 \times 3\).
- For the product \(BA\): Order of \(B\) is \(4 \times 3\) and order of \(A\) is \(3 \times 4\). Since inner dimensions match (3 = 3), the order of \(BA\) is \(4 \times 4\).
In simple words: Inner numbers must match for multiplication, and the outer numbers give the order of the resulting matrix.
Exam Tip: Matrix multiplication is not commutative; \(AB\) and \(BA\) can have completely different orders.

Question. If \(A = \begin{bmatrix} 3 & 1 \\ 2 & -3 \end{bmatrix} \), then find \(|\text{adj} A|\).
Answer: First, find the determinant of \(A\):
\(|A| = (3)(-3) - (1)(2) = -9 - 2 = -11\)
Using the formula for a \(2 \times 2\) matrix, \(|\text{adj} A| = |A|^{n-1} = |A|^{2-1} = |A|\):
\(|\text{adj} A| = -11\).
In simple words: For a \(2 \times 2\) matrix, the determinant of its adjoint is simply equal to the determinant of the matrix itself.
Exam Tip: Remember that \(|\text{adj} A| = |A|\) holds true specifically for \(2 \times 2\) matrices.

Question. Write \(A^{-1}\) for \(A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \).
Answer: Inverse formula is \(A^{-1} = \frac{1}{|A|} \text{adj} A\).
1. Find determinant \(|A|\):
\(|A| = (2)(3) - (5)(1) = 6 - 5 = 1\)
2. Find adjoint of \(A\):
\(\text{adj} A = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)
3. Calculate \(A^{-1}\):
\(A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \).
In simple words: Divide the adjoint matrix by the determinant value to get the inverse.
Exam Tip: Always verify that \(|A| \neq 0\) before attempting to find the inverse.

Question. If a matrix has 5 elements, write all possible orders it can have.
Answer: Since 5 is a prime number, its only factors are 1 and 5.
Therefore, the possible orders are \(1 \times 5\) and \(5 \times 1\).
In simple words: Find all pairs of integers whose product equals 5.
Exam Tip: The number of elements in a matrix of order \(m \times n\) is always equal to \(m \times n\).

Question. Write the value of \(x - y + z\) from the following equation: \(\begin{bmatrix} x + y + z \\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix} \)
Answer: From the given equations:
1) \(x + y + z = 9\)
2) \(x + z = 5\)
3) \(y + z = 7\)
Substitute equation (2) into equation (1):
\((x + z) + y = 9 \implies 5 + y = 9 \implies y = 4\)
Substitute \(y = 4\) into equation (3):
\(4 + z = 7 \implies z = 3\)
Substitute \(z = 3\) into equation (2):
\(x + 3 = 5 \implies x = 2\)
Now calculate \(x - y + z\):
\(x - y + z = 2 - 4 + 3 = 1\).
In simple words: Solve for individual variables by sequential substitution, then evaluate the requested expression.
Exam Tip: Check your final values of \(x, y, z\) by substituting them back into all original equations.

Question. Simplify: \(\cos\theta \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} \)
Answer: Multiply each matrix by its scalar multiplier:
\(= \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} \)
Add the corresponding elements:
\(= \begin{bmatrix} \cos^2\theta + \sin^2\theta & \sin\theta\cos\theta - \sin\theta\cos\theta \\ -\sin\theta\cos\theta + \sin\theta\cos\theta & \cos^2\theta + \sin^2\theta \end{bmatrix} \)
Using trigonometric identity \(\cos^2\theta + \sin^2\theta = 1\):
\(= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
In simple words: Distribute the trigonometric functions inside the matrices and combine terms using basic identities.
Exam Tip: Look out for complementary cancellation of terms like \(\sin\theta\cos\theta - \sin\theta\cos\theta\).

Question. If \(A^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \) and \(B = \begin{bmatrix} -1 & 2 & 1 \\ 3 & 2 & 3 \end{bmatrix} \), find \(A^T - B^T\).
Answer: First, find \(B^T\) by transposing matrix \(B\) (which is of order \(2 \times 3\), so \(B^T\) is of order \(3 \times 2\)):
\(B^T = \begin{bmatrix} -1 & 3 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \)
Now subtract \(B^T\) from \(A^T\):
\(A^T - B^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 3 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \)
\(= \begin{bmatrix} 3 - (-1) & 4 - 3 \\ -1 - 2 & 2 - 2 \\ 0 - 1 & 1 - 3 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}\).
In simple words: Transpose matrix \(B\) first, then subtract corresponding entries from \(A^T\).
Exam Tip: Ensure both matrices have identical dimensions before performing subtraction.

Question. If \(3A - B = \begin{bmatrix} 5 & 0 \\ 1 & 1 \end{bmatrix} \) and \(B = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \), find the value of matrix \(B\) (Note: usually finds matrix \(A\)).
Answer: To find matrix \(A\):
\(3A = \begin{bmatrix} 5 & 0 \\ 1 & 1 \end{bmatrix} + B \)
Substitute matrix \(B\):
\(3A = \begin{bmatrix} 5 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \)
\(3A = \begin{bmatrix} 5 + 4 & 0 + 3 \\ 1 + 2 & 1 + 5 \end{bmatrix} = \begin{bmatrix} 9 & 3 \\ 3 & 6 \end{bmatrix} \)
Divide by 3 to find \(A\):
\(A = \frac{1}{3} \begin{bmatrix} 9 & 3 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} \).
In simple words: Add matrix \(B\) to the right side and divide every element by 3 to isolate \(A\).
Exam Tip: Read carefully whether the question asks for matrix \(A\) or matrix \(B\).

Question. If matrix \(A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \(A^2 = kA\), then write the value of \(k\).
Answer: Calculate \(A^2 = A \cdot A\):
\(A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\(= \begin{bmatrix} (1)(1) + (-1)(-1) & (1)(-1) + (-1)(1) \\ (-1)(1) + (1)(-1) & (-1)(-1) + (1)(1) \end{bmatrix}\)
\(= \begin{bmatrix} 1 + 1 & -1 - 1 \\ -1 - 1 & 1 + 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \)
Given \(A^2 = kA\):
\(\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \)
Comparing corresponding elements:
\(k = 2\).
In simple words: Square matrix \(A\) by multiplying it with itself, then compare it to \(kA\) to find \(k\).
Exam Tip: Checking any single corresponding element is enough to find scalar \(k\).

Question. If A is a square matrix of order 3 such that \(|\text{adj} A| = 81\), find \(|A|\).
Answer: Using the standard property: \(|\text{adj} A| = |A|^{n-1}\).
Given order \(n = 3\) and \(|\text{adj} A| = 81\):
\(|A|^{3-1} = 81\)
\(|A|^2 = 81\)
\(|A| = \pm 9\).
In simple words: Take the square root of the adjoint determinant since the order is 3.
Exam Tip: Remember both positive and negative values when extracting square roots for determinants.

Question. If A and B are symmetric matrices of same order, under what condition can you say that \(AB\) is also symmetric?
Answer: For \(AB\) to be symmetric, we must have \((AB)^T = AB\).
We know that \((AB)^T = B^T A^T\).
Since \(A\) and \(B\) are symmetric matrices, \(A^T = A\) and \(B^T = B\).
Thus, \((AB)^T = BA\).
Therefore, for \(AB\) to be symmetric, we need \(BA = AB\).
In simple words: The product of two symmetric matrices is symmetric only if they commute with each other (\(AB = BA\)).
Exam Tip: This is a classic conceptual result frequently tested in board examinations.