Q 1: Find the adjoint of the matrix \begin{bmatrix}1&2\\3&4\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}1&2\\3&4\end{bmatrix}.
To find adj A, we first need to find the cofactor of each element.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij},
where M_{ij} is the minor of the element a_{ij}.
- Cofactor of a_{11} (1): A_{11}=(-1)^{1+1}(4)=4
- Cofactor of a_{12} (2): A_{12}=(-1)^{1+2}(3)=-3
- Cofactor of a_{21} (3): A_{21}=(-1)^{2+1}(2)=-2
- Cofactor of a_{22} (4): A_{22}=(-1)^{2+2}(1)=1
Writing these cofactors in matrix form, we get the Cofactor Matrix:
\begin{bmatrix}4&-3\\-2&1\end{bmatrix}
The adjoint of matrix A is simply the transpose of this cofactor matrix:
Final Answer: adj A=\begin{bmatrix}4&-2\\-3&1\end{bmatrix}
Teacher’s Pro-Tip: For a 2\times 2 matrix, you can use a quick shortcut!
Just interchange the diagonal elements (a_{11} and a_{22}) and change the signs of the off-diagonal elements (a_{12} and a_{21}) to get the adjoint instantly.
Q 2: Find the adjoint of the matrix \begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}
Solution:
Let A=\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}.
We need to calculate the cofactor A_{ij} for every single element in this 3\times 3 matrix.
Calculating Row 1 Cofactors:
- A_{11}=(-1)^{1+1}\begin{vmatrix}3&5\\0&1\end{vmatrix}=1\cdot(3(1)-5(0))=3-0=3
- A_{12}=(-1)^{1+2}\begin{vmatrix}2&5\\-2&1\end{vmatrix}=-1\cdot(2(1)-5(-2))=-(2+10)=-12
- A_{13}=(-1)^{1+3}\begin{vmatrix}2&3\\-2&0\end{vmatrix}=1\cdot(2(0)-3(-2))=0+6=6
Calculating Row 2 Cofactors:
- A_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-1\cdot(-1(1)-2(0))=-(-1-0)=1
- A_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\-2&1\end{vmatrix}=1\cdot(1(1)-2(-2))=1+4=5
- A_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\-2&0\end{vmatrix}=-1\cdot(1(0)-(-1)(-2))=-(0-2)=2
Calculating Row 3 Cofactors:
- A_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\3&5\end{vmatrix}=1\cdot(-1(5)-2(3))=-5-6=-11
- A_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\2&5\end{vmatrix}=-1\cdot(1(5)-2(2))=-(5-4)=-1
- A_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\2&3\end{vmatrix}=1\cdot(1(3)-(-1)(2))=3+2=5
Now, we arrange these into the Cofactor Matrix:
\begin{bmatrix}3&-12&6\\1&5&2\\-11&-1&5\end{bmatrix}
Taking the transpose of the cofactor matrix gives us the final adj A:
Final Answer: adj A=\begin{bmatrix}3&1&-11\\-12&5&-1\\6&2&5\end{bmatrix}
Q 3: Verify A(adj A) = (adj A)A = |A|I for the matrix \begin{bmatrix}2&3\\-4&-6\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}2&3\\-4&-6\end{bmatrix}.
To verify this, we first need to find the determinant |A| and the cofactor of each element to find adj A.
|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}, where M_{ij} is the minor of the element a_{ij}.
- Cofactor of a_{11} (2): A_{11}=(-1)^{1+1}(-6)=-6
- Cofactor of a_{12} (3): A_{12}=(-1)^{1+2}(-4)=4
- Cofactor of a_{21} (-4): A_{21}=(-1)^{2+1}(3)=-3
- Cofactor of a_{22} (-6): A_{22}=(-1)^{2+2}(2)=2
Writing these cofactors in matrix form, we get the Cofactor Matrix:
\begin{bmatrix}-6&4\\-3&2\end{bmatrix}
The adjoint of matrix A is simply the transpose of this cofactor matrix:
adj A=\begin{bmatrix}-6&-3\\4&2\end{bmatrix}
Now, let’s verify the equation:
A(adj A) = \begin{bmatrix}2&3\\-4&-6\end{bmatrix}\begin{bmatrix}-6&-3\\4&2\end{bmatrix} = \begin{bmatrix}-12+12&-6+6\\24-24&12-12\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}
(adj A)A = \begin{bmatrix}-6&-3\\4&2\end{bmatrix}\begin{bmatrix}2&3\\-4&-6\end{bmatrix} = \begin{bmatrix}-12+12&-18+18\\8-8&12-12\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}
|A|I = 0 \begin{bmatrix}1&0\\0&1\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}
Final Answer: A(adj A) = (adj A)A = |A|I = \begin{bmatrix}0&0\\0&0\end{bmatrix}. Hence verified.
Q 4: Verify A(adj A) = (adj A)A = |A|I for the matrix \begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}.
First, find the determinant |A|:
|A| = 1(0 - 0) - (-1)(9 - (-2)) + 2(0 - 0) = 0 + 11 + 0 = 11.
Next, find the cofactor A_{ij} = (-1)^{i+j}M_{ij} for each element.
- A_{11} = + (0 - 0) = 0
- A_{12} = - (9 - (-2)) = -11
- A_{13} = + (0 - 0) = 0
- A_{21} = - (-3 - 0) = 3
- A_{22} = + (3 - 2) = 1
- A_{23} = - (0 - (-1)) = -1
- A_{31} = + (2 - 0) = 2
- A_{32} = - (-2 - 6) = 8
- A_{33} = + (0 - (-3)) = 3
Writing these cofactors in matrix form gives the Cofactor Matrix:
\begin{bmatrix}0&-11&0\\3&1&-1\\2&8&3\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}
Now, verify the equation:
A(adj A) = \begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix} = \begin{bmatrix}0+11+0&3-1-2&2-8+6\\0+0+0&9+0+2&6+0-6\\0+0+0&3+0-3&2+0+9\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}
(adj A)A = \begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix} = \begin{bmatrix}0+9+2&0+0+0&0-6+6\\-11+3+8&11+0+0&-22-2+24\\0-3+3&0+0+0&0+2+9\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}
|A|I = 11 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}
Final Answer: A(adj A) = (adj A)A = |A|I. Hence verified.
Q 5: Find the inverse of the matrix \begin{bmatrix}2&-2\\4&3\end{bmatrix}
Solution: Let the given matrix be A=\begin{bmatrix}2&-2\\4&3\end{bmatrix}.
To find A^{-1}, we use the formula A^{-1} = \frac{1}{|A|}(adj A).
First, find the determinant |A| = (2)(3) - (-2)(4) = 6 + 8 = 14.
Since |A| [cite_start]\neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11}=(-1)^{1+1}(3)=3
- A_{12}=(-1)^{1+2}(4)=-4
- A_{21}=(-1)^{2+1}(-2)=2
- A_{22}=(-1)^{2+2}(2)=2
Cofactor Matrix:
\begin{bmatrix}3&-4\\2&2\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}3&2\\-4&2\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}
Q 6: Find the inverse of the matrix \begin{bmatrix}-1&5\\-3&2\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}-1&5\\-3&2\end{bmatrix}.
Determinant |A| = (-1)(2) - (5)(-3) = -2 + 15 = 13.
Since |A| [cite_start]\neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11}=(-1)^{1+1}(2)=2
- A_{12}=(-1)^{1+2}(-3)=3
- A_{21}=(-1)^{2+1}(5)=-5
- A_{22}=(-1)^{2+2}(-1)=-1
Cofactor Matrix:
\begin{bmatrix}2&3\\-5&-1\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}2&-5\\3&-1\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{13}\begin{bmatrix}2&-5\\3&-1\end{bmatrix}
Q 7: Find the inverse of the matrix \begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}.
Determinant |A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10.
Since |A| \neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11} = + (10 - 0) = 10
- A_{12} = - (0 - 0) = 0
- A_{13} = + (0 - 0) = 0
- A_{21} = - (10 - 0) = -10
- A_{22} = + (5 - 0) = 5
- A_{23} = - (0 - 0) = 0
- A_{31} = + (8 - 6) = 2
- A_{32} = - (4 - 0) = -4
- A_{33} = + (2 - 0) = 2
Cofactor Matrix:
\begin{bmatrix}10&0&0\\-10&5&0\\2&-4&2\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}10&-10&2\\0&5&-4\\0&0&2\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{10}\begin{bmatrix}10&-10&2\\0&5&-4\\0&0&2\end{bmatrix}
Q 8: Find the inverse of the matrix \begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}.
Determinant |A| = 1(-3 - 0) - 0 + 0 = -3.
Since |A| \neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11} = + (-3 - 0) = -3
- A_{12} = - (-3 - 0) = 3
- A_{13} = + (6 - 15) = -9
- A_{21} = - (0 - 0) = 0
- A_{22} = + (-1 - 0) = -1
- A_{23} = - (2 - 0) = -2
- A_{31} = + (0 - 0) = 0
- A_{32} = - (0 - 0) = 0
- A_{33} = + (3 - 0) = 3
Cofactor Matrix:
\begin{bmatrix}-3&3&-9\\0&-1&-2\\0&0&3\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}
Final Answer: A^{-1} = -\frac{1}{3}\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}
Q 9: Find the inverse of the matrix \begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}.
Determinant |A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7) = -2 - 4 + 3 = -3.
Since |A| \neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11} = + (-1 - 0) = -1
- A_{12} = - (4 - 0) = -4
- A_{13} = + (8 - 7) = 1
- A_{21} = - (1 - 6) = 5
- A_{22} = + (2 - (-21)) = 23
- A_{23} = - (4 - (-7)) = -11
- A_{31} = + (0 - (-3)) = 3
- A_{32} = - (0 - 12) = 12
- A_{33} = + (-2 - 4) = -6
Cofactor Matrix:
\begin{bmatrix}-1&-4&1\\5&23&-11\\3&12&-6\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}
Final Answer: A^{-1} = -\frac{1}{3}\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}
Q 10: Find the inverse of the matrix \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}.
Determinant |A| = 1(8 - 6) - (-1)(0 - (-9)) + 2(0 - 6) = 2 + 9 - 12 = -1.
Since |A| \neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11} = + (8 - 6) = 2
- A_{12} = - (0 - (-9)) = -9
- A_{13} = + (0 - 6) = -6
- A_{21} = - (-4 - (-4)) = 0
- A_{22} = + (4 - 6) = -2
- A_{23} = - (-2 - (-3)) = -1
- A_{31} = + (3 - 4) = -1
- A_{32} = - (-3 - 0) = 3
- A_{33} = + (2 - 0) = 2
Cofactor Matrix:
\begin{bmatrix}2&-9&-6\\0&-2&-1\\-1&3&2\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}
Since |A| = -1, the inverse is exactly the negative of the adjoint matrix.
Final Answer: A^{-1} = \frac{1}{-1}\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix} = \begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
Teacher’s Pro-Tip: A matrix is invertible if and only if it is a nonsingular matrix (i.e., |A| \neq 0).
Always double-check your sign configurations when generating your cofactor matrix for 3\times3 matrices; it’s the easiest place to drop a negative sign!
Q 11: Find the inverse of the matrix \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}
Solution:
Let the given matrix be A=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}.
First, find the determinant |A|:
|A| = 1(-\cos^2\alpha - \sin^2\alpha) - 0 + 0 = -(\cos^2\alpha + \sin^2\alpha) = -1.
Since |A| \neq 0, the inverse exists.
The cofactor A_{ij} is calculated as (-1)^{i+j}M_{ij}:
- A_{11} = + (-\cos^2\alpha - \sin^2\alpha) = -1
- A_{12} = - (0 - 0) = 0
- A_{13} = + (0 - 0) = 0
- A_{21} = - (0 - 0) = 0
- A_{22} = + (-\cos\alpha - 0) = -\cos\alpha
- A_{23} = - (\sin\alpha - 0) = -\sin\alpha
- A_{31} = + (0 - 0) = 0
- A_{32} = - (\sin\alpha - 0) = -\sin\alpha
- A_{33} = + (\cos\alpha - 0) = \cos\alpha
Cofactor Matrix:
\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}
The adjoint of matrix A is the transpose of this cofactor matrix:
adj A=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{-1}\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix} = \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}
Q 12: Let A=\begin{bmatrix}3&7\\2&5\end{bmatrix} and B=\begin{bmatrix}6&8\\7&9\end{bmatrix}. Verify that (AB)^{-1} = B^{-1}A^{-1}.
Solution:
First, calculate AB:
AB = \begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix} = \begin{bmatrix}18+49 & 24+63\\12+35 & 16+45\end{bmatrix} = \begin{bmatrix}67&87\\47&61\end{bmatrix}
Determinant |AB| = (67)(61) - (87)(47) = 4087 - 4089 = -2.
(AB)^{-1} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} = \begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}
Now, find A^{-1} and B^{-1}:
|A| = 15 - 14 = 1 \implies A^{-1} = \begin{bmatrix}5&-7\\-2&3\end{bmatrix}
|B| = 54 - 56 = -2 \implies B^{-1} = \frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}
Multiply B^{-1}A^{-1}:
B^{-1}A^{-1} = \frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix} = \frac{1}{-2}\begin{bmatrix}45+16 & -63-24\\-35-12 & 49+18\end{bmatrix} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} = \begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}
Final Answer: (AB)^{-1} = B^{-1}A^{-1}. Hence verified.
Q 13: If A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}, show that A^2 - 5A + 7I = O. Hence find A^{-1}.
Solution:
Calculate A^2:
A^2 = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix} = \begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix} = \begin{bmatrix}8&5\\-5&3\end{bmatrix}
Substitute into the equation:
A^2 - 5A + 7I = \begin{bmatrix}8&5\\-5&3\end{bmatrix} - 5\begin{bmatrix}3&1\\-1&2\end{bmatrix} + 7\begin{bmatrix}1&0\\0&1\end{bmatrix}
= \begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix} = O
To find A^{-1}, multiply the equation by A^{-1}:
A^{-1}(A^2 - 5A + 7I) = A^{-1}O
A - 5I + 7A^{-1} = O \implies 7A^{-1} = 5I - A
A^{-1} = \frac{1}{7}\left( \begin{bmatrix}5&0\\0&5\end{bmatrix} - \begin{bmatrix}3&1\\-1&2\end{bmatrix} \right) = \frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}
Q 14: For the matrix A=\begin{bmatrix}3&2\\1&1\end{bmatrix}, find the numbers a and b such that A^2 + aA + bI = O.
Solution:
Calculate A^2:
A^2 = \begin{bmatrix}3&2\\1&1\end{bmatrix}\begin{bmatrix}3&2\\1&1\end{bmatrix} = \begin{bmatrix}9+2 & 6+2\\3+1 & 2+1\end{bmatrix} = \begin{bmatrix}11&8\\4&3\end{bmatrix}
Substitute into the equation A^2 + aA + bI = O:
\begin{bmatrix}11&8\\4&3\end{bmatrix} + a\begin{bmatrix}3&2\\1&1\end{bmatrix} + b\begin{bmatrix}1&0\\0&1\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}
\begin{bmatrix}11+3a+b & 8+2a\\4+a & 3+a+b\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}
Comparing elements:
4 + a = 0 \implies a = -4
11 + 3a + b = 0 \implies 11 + 3(-4) + b = 0 \implies 11 - 12 + b = 0 \implies b = 1
Final Answer: a = -4, b = 1
Q 15: For the matrix A=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}, Show that A^3 - 6A^2 + 5A + 11I = O. Hence, find A^{-1}.
Solution:
Calculate A^2:
A^2 = \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} = \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}
Calculate A^3:
A^3 = A^2 \cdot A = \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} = \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}
Substitute into A^3 - 6A^2 + 5A + 11I:
\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix} - \begin{bmatrix}24&12&6\\-18&48&-84\\42&-18&84\end{bmatrix} + \begin{bmatrix}5&5&5\\5&10&-15\\10&-5&15\end{bmatrix} + \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O
To find A^{-1}, multiply by A^{-1}:
A^2 - 6A + 5I + 11A^{-1} = O \implies 11A^{-1} = 6A - A^2 - 5I
11A^{-1} = \begin{bmatrix}6&6&6\\6&12&-18\\12&-6&18\end{bmatrix} - \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix} - \begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix} = \begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}
Q 16: If A=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}, Verify that A^3 - 6A^2 + 9A - 4I = O and hence find A^{-1}.
Solution:
Calculate A^2:
A^2 = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} = \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}
Calculate A^3:
A^3 = A^2 \cdot A = \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} = \begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}
Substitute into A^3 - 6A^2 + 9A - 4I:
\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix} - \begin{bmatrix}36&-30&30\\-30&36&-30\\30&-30&36\end{bmatrix} + \begin{bmatrix}18&-9&9\\-9&18&-9\\9&-9&18\end{bmatrix} - \begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O
To find A^{-1}, multiply by A^{-1}:
A^2 - 6A + 9I - 4A^{-1} = O \implies 4A^{-1} = A^2 - 6A + 9I
4A^{-1} = \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix} - \begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix} + \begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix} = \begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
Final Answer: A^{-1} = \frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
Q 17: Let A be a nonsingular square matrix of order 3\times3. Then |adj A| is equal to
(A) |A|
(B) |A|^2
(C) |A|^3
(D) 3|A|
Solution:
We know the property: A(adj A) = |A|I.
Taking the determinant on both sides for a matrix of order n:
|A(adj A)| = ||A|I| |A| |adj A| = |A|^n |I|
Since A is nonsingular (|A| \neq 0) and order n = 3:
|A| |adj A| = |A|^3(1) |adj A|= |A|^2.
Final Answer: (B) |A|^2
Q 18: If A is an invertible matrix of order 2, then \det(A^{-1}) is equal to
(A) \det(A)
(B) \frac{1}{\det(A)}
(C) 1
(D) 0
Solution:
For an invertible matrix A, we know:
AA^{-1} = I
Taking the determinant on both sides:
|AA^{-1}|= |I| |A| |A^{-1}|= 1
(since the determinant of a product is the product of determinants,
|AB| = |A||B|).
|A^{-1}| = \frac{1}{|A|}
Therefore, \det(A^{-1}) = \frac{1}{\det(A)}.
Final Answer: (B) \frac{1}{\det(A)}