Samacheer Kalvi Class 11 Business Maths Solutions Chapter 1 Matrices and Determinants Exercise 1.2

Question 1. Find the adjoint of the matrix \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \).
Answer: The given matrix is \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \).
For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint is found by swapping the diagonal elements (a and d) and changing the signs of the off-diagonal elements (b and c).
\( \implies \) So, for \( A \), the adjoint is \( \text{Adj } A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \).
In simple words: To find the adjoint of a 2x2 matrix, just switch the numbers on the main diagonal and change the signs of the other two numbers.

🎯 Exam Tip: Remember the shortcut for the adjoint of a 2x2 matrix as it saves time. For larger matrices, you'll need to use cofactors.

 

Question 2. If \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \) then verify that \( A(\text{adj } A) = |A| I \) and also find \( A^{-1} \).
Answer: The given matrix is \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \).
First, we calculate the determinant of A, denoted as \( |A| \):
\( |A| = 1 \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} \)
\( \implies |A| = 1(4 \times 4 - 3 \times 3) - 3(1 \times 4 - 3 \times 1) + 3(1 \times 3 - 4 \times 1) \)
\( \implies |A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) \)
\( \implies |A| = 1(7) - 3(1) + 3(-1) \)
\( \implies |A| = 7 - 3 - 3 \)
\( \implies |A| = 1 \). Since \( |A| \neq 0 \), the inverse exists.
Next, we find the cofactor matrix \( [A_{ij}] \). The cofactor \( C_{ij} \) is \( (-1)^{i+j} \) times the determinant of the sub-matrix obtained by removing row i and column j.

Then, we find the adjoint of A, which is the transpose of the cofactor matrix.
\( \text{adj } A = [A_{ij}]^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \)
Now, we verify \( A(\text{adj } A) = |A| I \).
\( A(\text{adj } A) = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \)
\( \implies A(\text{adj } A) = \begin{bmatrix} (1 \cdot 7 + 3 \cdot (-1) + 3 \cdot (-1)) & (1 \cdot (-3) + 3 \cdot 1 + 3 \cdot 0) & (1 \cdot (-3) + 3 \cdot 0 + 3 \cdot 1) \\ (1 \cdot 7 + 4 \cdot (-1) + 3 \cdot (-1)) & (1 \cdot (-3) + 4 \cdot 1 + 3 \cdot 0) & (1 \cdot (-3) + 4 \cdot 0 + 3 \cdot 1) \\ (1 \cdot 7 + 3 \cdot (-1) + 4 \cdot (-1)) & (1 \cdot (-3) + 3 \cdot 1 + 4 \cdot 0) & (1 \cdot (-3) + 3 \cdot 0 + 4 \cdot 1) \end{bmatrix} \)
\( \implies A(\text{adj } A) = \begin{bmatrix} (7 - 3 - 3) & (-3 + 3 + 0) & (-3 + 0 + 3) \\ (7 - 4 - 3) & (-3 + 4 + 0) & (-3 + 0 + 3) \\ (7 - 3 - 4) & (-3 + 3 + 0) & (-3 + 0 + 4) \end{bmatrix} \)
\( \implies A(\text{adj } A) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{ ... (1)} \)
Now, calculate \( |A| I \). Since \( |A| = 1 \) and \( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \):
\( |A| I = 1 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{ ... (2)} \)
From (1) and (2), we see that \( A(\text{adj } A) = |A| I \), which verifies the property.
Finally, we find the inverse \( A^{-1} \). The formula for the inverse is \( A^{-1} = \frac{1}{|A|} (\text{adj } A) \).
Since \( |A| = 1 \):
\( A^{-1} = \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \)
\( \implies A^{-1} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \). This calculation is vital in many areas of linear algebra.
In simple words: First, we calculated a special number for the matrix called the determinant. Then we found another matrix called the adjoint. We checked that when you multiply the original matrix by its adjoint, you get the determinant times an identity matrix (a matrix with 1s on the diagonal and 0s everywhere else). Finally, we used the determinant and adjoint to find the inverse matrix.

🎯 Exam Tip: Always check if the determinant is non-zero before attempting to find the inverse, as a matrix with a zero determinant has no inverse.

 

Question 3. Find the inverse of each of the following matrices:
(i) \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 3 & 1 \\ -1 & 3 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix} \)
(iv) \( \begin{bmatrix} -3 & -5 & 4 \\ -2 & 3 & -1 \\ 1 & -4 & -6 \end{bmatrix} \)
Answer:
(i) Let \( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \).
First, find the determinant: \( |A| = (1 \times 3) - (-1 \times 2) = 3 - (-2) = 3 + 2 = 5 \).
Next, find the adjoint: \( \text{adj } A = \begin{bmatrix} 3 & -(-1) \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix} \).
Finally, find the inverse: \( A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{5} \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix} \). This method for a 2x2 matrix is quite straightforward.

(ii) Let \( A = \begin{bmatrix} 3 & 1 \\ -1 & 3 \end{bmatrix} \).
First, find the determinant: \( |A| = (3 \times 3) - (1 \times (-1)) = 9 - (-1) = 9 + 1 = 10 \).
Next, find the adjoint: \( \text{adj } A = \begin{bmatrix} 3 & -1 \\ -(-1) & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 1 & 3 \end{bmatrix} \).
Finally, find the inverse: \( A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{10} \begin{bmatrix} 3 & -1 \\ 1 & 3 \end{bmatrix} \).

(iii) Let \( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix} \).
First, find the determinant: For an upper triangular matrix like this, the determinant is the product of its diagonal elements.
\( |A| = 1 \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} - 2 \begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} + 3 \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} \)
\( \implies |A| = 1(2 \times 5 - 4 \times 0) - 2(0 \times 5 - 4 \times 0) + 3(0 \times 0 - 2 \times 0) \)
\( \implies |A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) \)
\( \implies |A| = 10 - 0 + 0 = 10 \).
Next, find the cofactor matrix \( [A_{ij}] \):

Find the adjoint: \( \text{adj } A = [A_{ij}]^T = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \).
Finally, find the inverse: \( A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \).

(iv) Let \( A = \begin{bmatrix} -3 & -5 & 4 \\ -2 & 3 & -1 \\ 1 & -4 & -6 \end{bmatrix} \).
First, find the determinant: \( |A| = -3 \begin{vmatrix} 3 & -1 \\ -4 & -6 \end{vmatrix} - (-5) \begin{vmatrix} -2 & -1 \\ 1 & -6 \end{vmatrix} + 4 \begin{vmatrix} -2 & 3 \\ 1 & -4 \end{vmatrix} \)
\( \implies |A| = -3((3)(-6) - (-1)(-4)) + 5((-2)(-6) - (-1)(1)) + 4((-2)(-4) - (3)(1)) \)
\( \implies |A| = -3(-18 - 4) + 5(12 + 1) + 4(8 - 3) \)
\( \implies |A| = -3(-22) + 5(13) + 4(5) \)
\( \implies |A| = 66 + 65 + 20 = 151 \).
Next, find the cofactor matrix \( [A_{ij}] \):

Find the adjoint: \( \text{adj } A = [A_{ij}]^T = \begin{bmatrix} -22 & -46 & -7 \\ -13 & 14 & 11 \\ 5 & -17 & -19 \end{bmatrix} \).
Finally, find the inverse: \( A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{151} \begin{bmatrix} -22 & -46 & -7 \\ -13 & 14 & 11 \\ 5 & -17 & -19 \end{bmatrix} \). This process of finding the inverse is a cornerstone of matrix algebra.
In simple words: For each matrix, we first calculated its determinant. If the determinant was not zero, we then found the adjoint matrix (which is the transpose of the cofactor matrix). Finally, we divided the adjoint matrix by the determinant to get the inverse matrix.

🎯 Exam Tip: Be very careful with signs when calculating cofactors, especially the \( (-1)^{i+j} \) part. A single sign error can lead to a completely wrong inverse.

 

Question 4. If \( A = \begin{bmatrix} 2 & 3 \\ 1 & -6 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 4 \\ 1 & -2 \end{bmatrix} \), then verify \( \text{adj}(AB) = (\text{adj } B)(\text{adj } A) \).
Answer: We are given two matrices: \( A = \begin{bmatrix} 2 & 3 \\ 1 & -6 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 4 \\ 1 & -2 \end{bmatrix} \).
First, calculate the product \( AB \):
\( AB = \begin{bmatrix} 2 & 3 \\ 1 & -6 \end{bmatrix} \begin{bmatrix} -1 & 4 \\ 1 & -2 \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} (2 \cdot (-1) + 3 \cdot 1) & (2 \cdot 4 + 3 \cdot (-2)) \\ (1 \cdot (-1) + (-6) \cdot 1) & (1 \cdot 4 + (-6) \cdot (-2)) \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} (-2 + 3) & (8 - 6) \\ (-1 - 6) & (4 + 12) \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} 1 & 2 \\ -7 & 16 \end{bmatrix} \).
Next, find \( \text{adj}(AB) \). For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint is \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
\( \text{adj}(AB) = \begin{bmatrix} 16 & -2 \\ -(-7) & 1 \end{bmatrix} = \begin{bmatrix} 16 & -2 \\ 7 & 1 \end{bmatrix} \text{ ... (1)} \).
Now, we calculate \( (\text{adj } B)(\text{adj } A) \).
First, find \( \text{adj } B \): \( \text{adj } B = \begin{bmatrix} -2 & -4 \\ -1 & -1 \end{bmatrix} \).
Next, find \( \text{adj } A \): \( \text{adj } A = \begin{bmatrix} -6 & -3 \\ -1 & 2 \end{bmatrix} \).
Now, calculate the product \( (\text{adj } B)(\text{adj } A) \):
\( (\text{adj } B)(\text{adj } A) = \begin{bmatrix} -2 & -4 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ -1 & 2 \end{bmatrix} \)
\( \implies (\text{adj } B)(\text{adj } A) = \begin{bmatrix} ((-2) \cdot (-6) + (-4) \cdot (-1)) & ((-2) \cdot (-3) + (-4) \cdot 2) \\ ((-1) \cdot (-6) + (-1) \cdot (-1)) & ((-1) \cdot (-3) + (-1) \cdot 2) \end{bmatrix} \)
\( \implies (\text{adj } B)(\text{adj } A) = \begin{bmatrix} (12 + 4) & (6 - 8) \\ (6 + 1) & (3 - 2) \end{bmatrix} \)
\( \implies (\text{adj } B)(\text{adj } A) = \begin{bmatrix} 16 & -2 \\ 7 & 1 \end{bmatrix} \text{ ... (2)} \).
From (1) and (2), we can clearly see that \( \text{adj}(AB) = (\text{adj } B)(\text{adj } A) \). This property is an important part of matrix algebra, especially when dealing with products of matrices.
In simple words: We multiplied matrix A by matrix B first, then found the adjoint of the result. Separately, we found the adjoint of B and the adjoint of A, then multiplied them in reverse order (adjoint B times adjoint A). We showed that both calculations give the same final matrix.

🎯 Exam Tip: Remember the property \( \text{adj}(AB) = (\text{adj } B)(\text{adj } A) \). Note the reversal of order for matrices B and A on the right side of the equation. This is similar to the inverse of a product, \( (AB)^{-1} = B^{-1}A^{-1} \).

 

Question 5. If \( A = \begin{bmatrix} 2 & -2 & 2 \\ 2 & 3 & 0 \\ 9 & 1 & 5 \end{bmatrix} \) then, show that \( (\text{adj } A) A = 0 \).
Answer: The given matrix is \( A = \begin{bmatrix} 2 & -2 & 2 \\ 2 & 3 & 0 \\ 9 & 1 & 5 \end{bmatrix} \).
First, we need to find the cofactor matrix \( [A_{ij}] \). The cofactor \( C_{ij} \) is \( (-1)^{i+j} \) times the determinant of the sub-matrix obtained by removing row i and column j.

Next, find the adjoint matrix, which is the transpose of the cofactor matrix.
\( \text{adj } A = [A_{ij}]^T = \begin{bmatrix} 15 & 12 & -6 \\ -10 & -8 & 4 \\ -25 & -20 & 10 \end{bmatrix} \).
Now, we multiply \( (\text{adj } A) \) by \( A \):
\( (\text{adj } A) A = \begin{bmatrix} 15 & 12 & -6 \\ -10 & -8 & 4 \\ -25 & -20 & 10 \end{bmatrix} \begin{bmatrix} 2 & -2 & 2 \\ 2 & 3 & 0 \\ 9 & 1 & 5 \end{bmatrix} \)
\( \implies (\text{adj } A) A = \begin{bmatrix} (15 \cdot 2 + 12 \cdot 2 + (-6) \cdot 9) & (15 \cdot (-2) + 12 \cdot 3 + (-6) \cdot 1) & (15 \cdot 2 + 12 \cdot 0 + (-6) \cdot 5) \\ ((-10) \cdot 2 + (-8) \cdot 2 + 4 \cdot 9) & ((-10) \cdot (-2) + (-8) \cdot 3 + 4 \cdot 1) & ((-10) \cdot 2 + (-8) \cdot 0 + 4 \cdot 5) \\ ((-25) \cdot 2 + (-20) \cdot 2 + 10 \cdot 9) & ((-25) \cdot (-2) + (-20) \cdot 3 + 10 \cdot 1) & ((-25) \cdot 2 + (-20) \cdot 0 + 10 \cdot 5) \end{bmatrix} \)
\( \implies (\text{adj } A) A = \begin{bmatrix} (30 + 24 - 54) & (-30 + 36 - 6) & (30 + 0 - 30) \\ (-20 - 16 + 36) & (20 - 24 + 4) & (-20 + 0 + 20) \\ (-50 - 40 + 90) & (50 - 60 + 10) & (-50 + 0 + 50) \end{bmatrix} \)
\( \implies (\text{adj } A) A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \).
Thus, we have shown that \( (\text{adj } A) A = 0 \). This is a special case because for any invertible matrix, \( (\text{adj } A) A = |A|I \). If this product is zero, it implies that the determinant \( |A| \) must be zero for the identity matrix to remain.
In simple words: We found the adjoint matrix for A first. Then, we multiplied this adjoint matrix by the original matrix A. After doing the multiplication, all the numbers in the resulting matrix turned out to be zero, which means the product is the zero matrix.

🎯 Exam Tip: A key property states that \( A(\text{adj } A) = (\text{adj } A)A = |A|I \). If the result is the zero matrix, it implies that the determinant \( |A| \) must be zero. This is a good way to check your determinant calculation.

 

Question 6. If \( A = \begin{bmatrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{bmatrix} \) then, show that the inverse of A is A itself.
Answer: The given matrix is \( A = \begin{bmatrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{bmatrix} \).
First, find the determinant of A: \( |A| \).
\( |A| = -1 \begin{vmatrix} -3 & 4 \\ -4 & 5 \end{vmatrix} - 2 \begin{vmatrix} 4 & 4 \\ 4 & 5 \end{vmatrix} + (-2) \begin{vmatrix} 4 & -3 \\ 4 & -4 \end{vmatrix} \)
\( \implies |A| = -1((-3)(5) - (4)(-4)) - 2((4)(5) - (4)(4)) - 2((4)(-4) - (-3)(4)) \)
\( \implies |A| = -1(-15 + 16) - 2(20 - 16) - 2(-16 + 12) \)
\( \implies |A| = -1(1) - 2(4) - 2(-4) \)
\( \implies |A| = -1 - 8 + 8 \)
\( \implies |A| = -1 \).
Since \( |A| \neq 0 \), the inverse exists.
Next, find the cofactor matrix \( [A_{ij}] \):

Find the adjoint matrix, which is the transpose of the cofactor matrix.
\( \text{adj } A = [A_{ij}]^T = \begin{bmatrix} 1 & -2 & 2 \\ -4 & 3 & -4 \\ -4 & 4 & -5 \end{bmatrix} \).
Finally, find the inverse \( A^{-1} = \frac{1}{|A|} (\text{adj } A) \).
Since \( |A| = -1 \):
\( A^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -2 & 2 \\ -4 & 3 & -4 \\ -4 & 4 & -5 \end{bmatrix} \)
\( \implies A^{-1} = \begin{bmatrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{bmatrix} \).
Comparing \( A^{-1} \) with the original matrix \( A \), we see that \( A^{-1} = A \). This property means the matrix is its own inverse, a special characteristic for certain matrices.
In simple words: We calculated the determinant and the adjoint of the matrix A. Then, using these, we found the inverse matrix \( A^{-1} \). We observed that the inverse matrix we calculated was exactly the same as the original matrix A, proving that A is its own inverse.

🎯 Exam Tip: When a matrix is its own inverse, multiplying it by itself yields the identity matrix. This is a special characteristic of involutory matrices.

 

Question 7. If \( A^{-1} = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \) then, find A.
Answer: We are given \( A^{-1} = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \).
We know that the inverse of an inverse matrix is the original matrix itself, i.e., \( (A^{-1})^{-1} = A \).
So, we need to find the inverse of the given \( A^{-1} \). Let \( B = A^{-1} \). We will find \( B^{-1} \).
First, find the determinant of B: \( |B| \).
\( |B| = 1 \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} - 0 \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \)
\( \implies |B| = 1(1 \cdot 1 - (-1) \cdot (-1)) - 0 + 3(2 \cdot (-1) - 1 \cdot 1) \)
\( \implies |B| = 1(1 - 1) - 0 + 3(-2 - 1) \)
\( \implies |B| = 1(0) - 0 + 3(-3) \)
\( \implies |B| = 0 - 0 - 9 = -9 \).
Since \( |B| \neq 0 \), the inverse exists.
Next, find the cofactor matrix \( [B_{ij}] \):

Find the adjoint matrix \( \text{adj } B = [B_{ij}]^T = \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 7 \\ -3 & 1 & 1 \end{bmatrix} \).
Finally, find \( B^{-1} = \frac{1}{|B|} (\text{adj } B) \).
\( B^{-1} = \frac{1}{-9} \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 7 \\ -3 & 1 & 1 \end{bmatrix} \).
\( \implies B^{-1} = \begin{bmatrix} 0 & \frac{-3}{-9} & \frac{-3}{-9} \\ \frac{-3}{-9} & \frac{-2}{-9} & \frac{7}{-9} \\ \frac{-3}{-9} & \frac{1}{-9} & \frac{1}{-9} \end{bmatrix} \)
\( \implies B^{-1} = \begin{bmatrix} 0 & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{9} & \frac{-7}{9} \\ \frac{1}{3} & \frac{-1}{9} & \frac{-1}{9} \end{bmatrix} \).
Therefore, \( A = \begin{bmatrix} 0 & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{9} & \frac{-7}{9} \\ \frac{1}{3} & \frac{-1}{9} & \frac{-1}{9} \end{bmatrix} \). Finding the inverse of an inverse is a good check of understanding matrix properties.
In simple words: Since we know that finding the inverse of an inverse matrix gives us the original matrix back, we simply treated the given \( A^{-1} \) as a new matrix and found its inverse. This calculation gave us matrix A.

🎯 Exam Tip: Remember the fundamental property \( (A^{-1})^{-1} = A \). This means you just need to calculate the inverse of the given inverse matrix to find A.

 

Question 8. Show that the matrices \( A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} \frac{4}{5} & \frac{-2}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{3}{5} & \frac{-1}{5} \\ \frac{-1}{5} & \frac{-2}{5} & \frac{4}{5} \end{bmatrix} \) are inverses of each other.
Answer: To show that two matrices A and B are inverses of each other, we need to prove that their product is the identity matrix, i.e., \( AB = BA = I \).
First, let's calculate \( AB \). We can factor out \( \frac{1}{5} \) from matrix B for easier calculation.
\( B = \frac{1}{5} \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{bmatrix} \).
\( AB = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} \left( \frac{1}{5} \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{bmatrix} \right) \)
\( \implies AB = \frac{1}{5} \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{bmatrix} \)
\( \implies AB = \frac{1}{5} \begin{bmatrix} (2 \cdot 4 + 2 \cdot (-1) + 1 \cdot (-1)) & (2 \cdot (-2) + 2 \cdot 3 + 1 \cdot (-2)) & (2 \cdot (-1) + 2 \cdot (-1) + 1 \cdot 4) \\ (1 \cdot 4 + 3 \cdot (-1) + 1 \cdot (-1)) & (1 \cdot (-2) + 3 \cdot 3 + 1 \cdot (-2)) & (1 \cdot (-1) + 3 \cdot (-1) + 1 \cdot 4) \\ (1 \cdot 4 + 2 \cdot (-1) + 2 \cdot (-1)) & (1 \cdot (-2) + 2 \cdot 3 + 2 \cdot (-2)) & (1 \cdot (-1) + 2 \cdot (-1) + 2 \cdot 4) \end{bmatrix} \)
\( \implies AB = \frac{1}{5} \begin{bmatrix} (8 - 2 - 1) & (-4 + 6 - 2) & (-2 - 2 + 4) \\ (4 - 3 - 1) & (-2 + 9 - 2) & (-1 - 3 + 4) \\ (4 - 2 - 2) & (-2 + 6 - 4) & (-1 - 2 + 8) \end{bmatrix} \)
\( \implies AB = \frac{1}{5} \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \).
Next, let's calculate \( BA \).
\( BA = \left( \frac{1}{5} \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{bmatrix} \right) \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} \)
\( \implies BA = \frac{1}{5} \begin{bmatrix} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{bmatrix} \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} \)
\( \implies BA = \frac{1}{5} \begin{bmatrix} (4 \cdot 2 + (-2) \cdot 1 + (-1) \cdot 1) & (4 \cdot 2 + (-2) \cdot 3 + (-1) \cdot 2) & (4 \cdot 1 + (-2) \cdot 1 + (-1) \cdot 2) \\ ((-1) \cdot 2 + 3 \cdot 1 + (-1) \cdot 1) & ((-1) \cdot 2 + 3 \cdot 3 + (-1) \cdot 2) & ((-1) \cdot 1 + 3 \cdot 1 + (-1) \cdot 2) \\ ((-1) \cdot 2 + (-2) \cdot 1 + 4 \cdot 1) & ((-1) \cdot 2 + (-2) \cdot 3 + 4 \cdot 2) & ((-1) \cdot 1 + (-2) \cdot 1 + 4 \cdot 2) \end{bmatrix} \)
\( \implies BA = \frac{1}{5} \begin{bmatrix} (8 - 2 - 1) & (8 - 6 - 2) & (4 - 2 - 2) \\ (-2 + 3 - 1) & (-2 + 9 - 2) & (-1 + 3 - 2) \\ (-2 - 2 + 4) & (-2 - 6 + 8) & (-1 - 2 + 8) \end{bmatrix} \)
\( \implies BA = \frac{1}{5} \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)
\( \implies BA = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \).
Since \( AB = I \) and \( BA = I \), the matrices A and B are inverses of each other. This verification confirms the fundamental definition of inverse matrices.
In simple words: To show that two matrices are inverses, we need to multiply them in both possible orders. If both multiplications result in the identity matrix (a matrix with ones on the main diagonal and zeros elsewhere), then they are indeed inverses of each other. We performed these multiplications and found that both gave the identity matrix.

🎯 Exam Tip: When verifying if two matrices are inverses, you must check both \( AB=I \) and \( BA=I \). For square matrices, if one holds, the other usually does too, but it's good practice to show both.

 

Question 9. If \( A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \), then verify that \( (AB)^{-1} = B^{-1}A^{-1} \).
Answer: We are given two matrices: \( A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \).
First, we will calculate the left-hand side (LHS), \( (AB)^{-1} \).
Calculate \( AB \):
\( AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} (3 \cdot 6 + 7 \cdot 7) & (3 \cdot 8 + 7 \cdot 9) \\ (2 \cdot 6 + 5 \cdot 7) & (2 \cdot 8 + 5 \cdot 9) \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} (18 + 49) & (24 + 63) \\ (12 + 35) & (16 + 45) \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix} \).
Now, find the determinant of \( AB \):
\( |AB| = (67 \times 61) - (87 \times 47) \)
\( \implies |AB| = 4087 - 4089 = -2 \).
Next, find the adjoint of \( AB \):
\( \text{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \).
Finally, find \( (AB)^{-1} \):
\( (AB)^{-1} = \frac{1}{|AB|} (\text{adj}(AB)) = \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \text{ ... (1)} \).
Now, we will calculate the right-hand side (RHS), \( B^{-1}A^{-1} \).
First, find \( B^{-1} \).
\( |B| = (6 \times 9) - (8 \times 7) = 54 - 56 = -2 \).
\( \text{adj } B = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} \).
\( B^{-1} = \frac{1}{|B|} (\text{adj } B) = \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} \).
Next, find \( A^{-1} \).
\( |A| = (3 \times 5) - (7 \times 2) = 15 - 14 = 1 \).
\( \text{adj } A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \).
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \).
Now, calculate \( B^{-1}A^{-1} \):
\( B^{-1}A^{-1} = \left( \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} \right) \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)
\( \implies B^{-1}A^{-1} = \frac{1}{-2} \begin{bmatrix} (9 \cdot 5 + (-8) \cdot (-2)) & (9 \cdot (-7) + (-8) \cdot 3) \\ ((-7) \cdot 5 + 6 \cdot (-2)) & ((-7) \cdot (-7) + 6 \cdot 3) \end{bmatrix} \)
\( \implies B^{-1}A^{-1} = \frac{1}{-2} \begin{bmatrix} (45 + 16) & (-63 - 24) \\ (-35 - 12) & (49 + 18) \end{bmatrix} \)
\( \implies B^{-1}A^{-1} = \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \text{ ... (2)} \).
From (1) and (2), we see that \( (AB)^{-1} = B^{-1}A^{-1} \). This property is consistent and holds for all invertible square matrices of the same order.
In simple words: We first multiplied matrices A and B, then found the inverse of that combined matrix. Separately, we found the inverse of B and the inverse of A, then multiplied them together in the order \( B^{-1} \) then \( A^{-1} \). Both results were the same, showing that the inverse of a product is the product of the inverses in reverse order.

🎯 Exam Tip: This verification demonstrates a crucial property of inverses: the inverse of a product of matrices is the product of their inverses in reverse order. This is a common test question to check understanding of matrix algebra rules.

 

Question 10. If \( \begin{bmatrix} 1 & 1 & 3 \\ 2 & \lambda & 4 \\ 9 & 7 & 11 \end{bmatrix} \) has no inverse, then find \( \lambda \).
Answer: A matrix has no inverse if and only if its determinant is zero.
Let \( M = \begin{bmatrix} 1 & 1 & 3 \\ 2 & \lambda & 4 \\ 9 & 7 & 11 \end{bmatrix} \).
We set the determinant of M to zero: \( |M| = 0 \).
\( |M| = 1 \begin{vmatrix} \lambda & 4 \\ 7 & 11 \end{vmatrix} - 1 \begin{vmatrix} 2 & 4 \\ 9 & 11 \end{vmatrix} + 3 \begin{vmatrix} 2 & \lambda \\ 9 & 7 \end{vmatrix} = 0 \)
\( \implies 1(\lambda \cdot 11 - 4 \cdot 7) - 1(2 \cdot 11 - 4 \cdot 9) + 3(2 \cdot 7 - \lambda \cdot 9) = 0 \)
\( \implies 1(11\lambda - 28) - 1(22 - 36) + 3(14 - 9\lambda) = 0 \)
\( \implies 11\lambda - 28 - (-14) + 42 - 27\lambda = 0 \)
\( \implies 11\lambda - 28 + 14 + 42 - 27\lambda = 0 \)
\( \implies (11\lambda - 27\lambda) + (-28 + 14 + 42) = 0 \)
\( \implies -16\lambda + 28 = 0 \)
\( \implies -16\lambda = -28 \)
\( \implies \lambda = \frac{-28}{-16} \)
\( \implies \lambda = \frac{7}{4} \). For the matrix to be non-invertible, this specific value of lambda is required.
In simple words: A matrix does not have an inverse if its determinant is zero. So, we calculated the determinant of the given matrix and set it equal to zero. Solving this equation helped us find the value of \( \lambda \).

🎯 Exam Tip: Always remember that a square matrix has no inverse if and only if its determinant is zero. This is a fundamental concept in linear algebra for determining invertibility.

 

Question 11. If \( X = \begin{bmatrix} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{bmatrix} \) and \( Y = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & p & q \end{bmatrix} \) then, find p, q if \( Y = X^{-1} \).
Answer: We are given \( X = \begin{bmatrix} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{bmatrix} \) and \( Y = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & p & q \end{bmatrix} \).
We are told that \( Y = X^{-1} \). This means that \( XY = I \), where I is the identity matrix.
So, we multiply X and Y and set the result equal to the identity matrix \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
\( XY = \begin{bmatrix} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{bmatrix} \begin{bmatrix} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & p & q \end{bmatrix} \)
\( \implies XY = \begin{bmatrix} (8 \cdot 2 + (-1) \cdot 0 + (-3) \cdot 5) & (8 \cdot 1 + (-1) \cdot 2 + (-3) \cdot p) & (8 \cdot (-1) + (-1) \cdot 1 + (-3) \cdot q) \\ ((-5) \cdot 2 + 1 \cdot 0 + 2 \cdot 5) & ((-5) \cdot 1 + 1 \cdot 2 + 2 \cdot p) & ((-5) \cdot (-1) + 1 \cdot 1 + 2 \cdot q) \\ (10 \cdot 2 + (-1) \cdot 0 + (-4) \cdot 5) & (10 \cdot 1 + (-1) \cdot 2 + (-4) \cdot p) & (10 \cdot (-1) + (-1) \cdot 1 + (-4) \cdot q) \end{bmatrix} \)
\( \implies XY = \begin{bmatrix} (16 + 0 - 15) & (8 - 2 - 3p) & (-8 - 1 - 3q) \\ (-10 + 0 + 10) & (-5 + 2 + 2p) & (5 + 1 + 2q) \\ (20 + 0 - 20) & (10 - 2 - 4p) & (-10 - 1 - 4q) \end{bmatrix} \)
\( \implies XY = \begin{bmatrix} 1 & 6 - 3p & -9 - 3q \\ 0 & -3 + 2p & 6 + 2q \\ 0 & 8 - 4p & -11 - 4q \end{bmatrix} \).
Now, we equate this to the identity matrix I:
\( \begin{bmatrix} 1 & 6 - 3p & -9 - 3q \\ 0 & -3 + 2p & 6 + 2q \\ 0 & 8 - 4p & -11 - 4q \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
By comparing the elements of the matrices, we get equations for p and q.
Comparing the element in the first row, second column:
\( 6 - 3p = 0 \)
\( \implies 3p = 6 \)
\( \implies p = 2 \).
Comparing the element in the first row, third column:
\( -9 - 3q = 0 \)
\( \implies -3q = 9 \)
\( \implies q = -3 \).
We can also verify these values using other elements. For example, using the second row, second column:
\( -3 + 2p = 1 \)
\( -3 + 2(2) = 1 \)
\( -3 + 4 = 1 \), which is true.
Using the second row, third column:
\( 6 + 2q = 0 \)
\( 6 + 2(-3) = 0 \)
\( 6 - 6 = 0 \), which is true.
Using the third row, second column:
\( 8 - 4p = 0 \)
\( 8 - 4(2) = 0 \)
\( 8 - 8 = 0 \), which is true.
Using the third row, third column:
\( -11 - 4q = 1 \)
\( -11 - 4(-3) = 1 \)
\( -11 + 12 = 1 \), which is true.
Thus, the values are \( p = 2 \) and \( q = -3 \). This method shows how the property of inverse matrices can be used to solve for unknown variables within a matrix.
In simple words: We are told that matrix Y is the inverse of matrix X. This means if you multiply X by Y, you should get the identity matrix (which has ones on the main diagonal and zeros everywhere else). We multiplied X and Y, and then set the parts of the resulting matrix equal to the numbers in the identity matrix to find the values of p and q.

🎯 Exam Tip: When \( Y = X^{-1} \), the most direct approach to find unknown variables in Y is to use the property \( XY = I \). This converts the problem into a system of equations by comparing elements of the product matrix with the identity matrix.