Exercise – 4.5
1. Find the adjoint of each of the matrices:
Solution:
-
$\left[ {\begin{array}{*{20}{r}}1&2\\3&4\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&2\\3&4\end{array}} \right]$. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(4) = 4$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(2) = -2$
$A_{22} = (-1)^{2+2}(1) = 1$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}4&-2\\-3&1\end{array}} \right]$
2. Find the adjoint of the matrix:
Solution:
$\left[ {\begin{array}{*{20}{r}}1&-1&2\\2&3&-2\\-2&0&1\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&-1&2\\2&3&-2\\-2&0&1\end{array}} \right]$. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}3&-2\\0&1\end{array}} \right] = 3$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}2&-2\\-2&1\end{array}} \right] = -12$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}2&3\\-2&0\end{array}} \right] = 6$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}-1&2\\0&1\end{array}} \right] = 1$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&2\\-2&1\end{array}} \right] = 5$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&-1\\-2&0\end{array}} \right] = 2$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}-1&2\\3&-2\end{array}} \right] = -11$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&2\\2&-2\end{array}} \right] = -1$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&-1\\2&3\end{array}} \right] = 5$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}3&-11&0\\1&5&-1\\-2&-1&5\end{array}} \right]^T = \left[ {\begin{array}{*{20}{r}}3&1&-2\\-11&5&-1\\0&-1&5\end{array}} \right]$
3. Verify: $A(adj A) = (adj A)A = |A|I$:
Solution:
-
$\left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right]$
$|A| = -12 + 12 = 0$
Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(-6) = -6$
$A_{12} = (-1)^{1+2}(-4) = 4$
$A_{21} = (-1)^{2+1}(3) = -3$
$A_{22} = (-1)^{2+2}(2) = 2$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}-6&-3\\4&2\end{array}} \right]$
$A(adj A) = \left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}-6&-3\\4&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right]$
$(adj A)A = \left[ {\begin{array}{*{20}{r}}-6&-3\\4&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right]$
Since $|A| = 0$, $|A|I = O$
Hence, $A(adj A) = (adj A)A = |A|I$ -
$\left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right]$
Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}0&-2\\0&3\end{array}} \right] = 0$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}3&-2\\1&3\end{array}} \right] = -11$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}3&0\\1&0\end{array}} \right] = 0$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}-1&2\\0&3\end{array}} \right] = 3$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&2\\1&3\end{array}} \right] = 1$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&-1\\1&0\end{array}} \right] = -1$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}-1&2\\0&-2\end{array}} \right] = 2$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&2\\3&-2\end{array}} \right] = 8$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&-1\\3&0\end{array}} \right] = 3$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}0&3&2\\-11&1&8\\0&-1&3\end{array}} \right]$
$A(adj A) = \left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}0&3&2\\-11&1&8\\0&-1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}11&0&0\\0&11&0\\0&0&11\end{array}} \right] = 11I$
$(adj A)A = \left[ {\begin{array}{*{20}{r}}0&3&2\\-11&1&8\\0&-1&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}11&0&0\\0&11&0\\0&0&11\end{array}} \right] = 11I$
Also, $|A| = 11$
Hence, $A(adj A) = (adj A)A = |A|I$
4. Find the inverse of each of the matrices (if it exists):
Solution:
-
$\left[ {\begin{array}{*{20}{r}}2&-2\\4&3\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}2&-2\\4&3\end{array}} \right]$
$|A| = 6 + 8 = 14 \neq 0$
So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(3) = 3$
$A_{12} = (-1)^{1+2}(4) = -4$
$A_{21} = (-1)^{2+1}(-2) = 2$
$A_{22} = (-1)^{2+2}(2) = 2$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}3&2\\-4&2\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{14}\left[ {\begin{array}{*{20}{r}}3&2\\-4&2\end{array}} \right]$ -
$\left[ {\begin{array}{*{20}{r}}-1&5\\-3&2\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}-1&5\\-3&2\end{array}} \right]$
$|A| = -2 + 15 = 13 \neq 0$
So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(2) = 2$
$A_{12} = (-1)^{1+2}(-3) = 3$
$A_{21} = (-1)^{2+1}(5) = -5$
$A_{22} = (-1)^{2+2}(-1) = -1$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}2&-5\\3&-1\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{13}\left[ {\begin{array}{*{20}{r}}2&-5\\3&-1\end{array}} \right]$ -
$\left[ {\begin{array}{*{20}{r}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$
$|A| = 10 \neq 0$
So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}2&4\\0&5\end{array}} \right] = 10$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}0&4\\0&5\end{array}} \right] = 0$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}0&2\\0&0\end{array}} \right] = 0$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}2&3\\0&5\end{array}} \right] = -10$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&3\\0&5\end{array}} \right] = 5$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&2\\0&0\end{array}} \right] = 0$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}2&3\\2&4\end{array}} \right] = 2$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&3\\0&4\end{array}} \right] = -4$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&2\\0&2\end{array}} \right] = 2$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}10&-10&2\\0&5&-4\\0&0&2\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}10&-10&2\\0&5&-4\\0&0&2\end{array}} \right]$ -
$\left[ {\begin{array}{*{20}{r}}1&0&0\\3&3&0\\5&2&-1\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&0&0\\3&3&0\\5&2&-1\end{array}} \right]$
$|A| = -3 – 0 = -3 \neq 0$
So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}3&0\\2&-1\end{array}} \right] = -3$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}3&0\\5&-1\end{array}} \right] = 3$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}3&3\\5&2\end{array}} \right] = -9$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}0&0\\2&-1\end{array}} \right] = 0$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&0\\5&-1\end{array}} \right] = -1$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&0\\5&2\end{array}} \right] = -2$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}0&0\\3&0\end{array}} \right] = 0$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&0\\3&0\end{array}} \right] = 0$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&0\\3&3\end{array}} \right] = 3$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}-3&0&0\\3&-1&0\\-9&-2&3\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{3}\left[ {\begin{array}{*{20}{r}}-3&0&0\\3&-1&0\\-9&-2&3\end{array}} \right]$ -
$\left[ {\begin{array}{*{20}{r}}2&1&3\\4&-1&0\\-7&2&1\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}2&1&3\\4&-1&0\\-7&2&1\end{array}} \right]$
$|A| = -3 \neq 0$
So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}-1&0\\2&1\end{array}} \right] = -1$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}4&0\\-7&1\end{array}} \right] = -4$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}4&-1\\-7&2\end{array}} \right] = 1$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}1&3\\2&1\end{array}} \right] = 5$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}2&3\\-7&1\end{array}} \right] = 23$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}2&1\\-7&2\end{array}} \right] = -11$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}1&3\\-1&0\end{array}} \right] = 3$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}2&3\\4&0\end{array}} \right] = 12$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}2&1\\4&-1\end{array}} \right] = -6$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}-1&-9&-6\\5&-2&-1\\3&12&2\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{3}\left[ {\begin{array}{*{20}{r}}-1&-9&-6\\5&-2&-1\\3&12&2\end{array}} \right]$ -
$\left[ {\begin{array}{*{20}{r}}1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha\end{array}} \right]$
$|A| = -(\cos^2\alpha + \sin^2\alpha) = -1 \neq 0$
So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}\cos \alpha&\sin \alpha\\\sin \alpha&-\cos \alpha\end{array}} \right] = -1$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}0&\sin \alpha\\0&-\cos \alpha\end{array}} \right] = 0$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}0&\cos \alpha\\0&\sin \alpha\end{array}} \right] = 0$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}0&0\\\sin \alpha&-\cos \alpha\end{array}} \right] = 0$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&0\\0&-\cos \alpha\end{array}} \right] = -\cos \alpha$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&0\\0&\sin \alpha\end{array}} \right] = -\sin \alpha$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}0&0\\\cos \alpha&\sin \alpha\end{array}} \right] = 0$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&0\\0&\sin \alpha\end{array}} \right] = -\sin \alpha$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&0\\0&\cos \alpha\end{array}} \right] = \cos \alpha$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}-1&0&0\\0&-\cos \alpha&-\sin \alpha\\0&-\sin \alpha&\cos \alpha\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = -1 \cdot \left[ {\begin{array}{*{20}{r}}-1&0&0\\0&-\cos \alpha&-\sin \alpha\\0&-\sin \alpha&\cos \alpha\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha\end{array}} \right]$
5. Let $A = \left[ {\begin{array}{*{20}{r}}3&7\\2&5\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{r}}6&8\\7&9\end{array}} \right]$. Verify that $(AB)^{-1} = B^{-1}A^{-1}$.
Solution:
We have, $AB = \left[ {\begin{array}{*{20}{r}}3&7\\2&5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}6&8\\7&9\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}67&87\\47&61\end{array}} \right]$
Since, $|AB| = 67 \times 61 – 47 \times 87 = -2 \neq 0$
So, AB is a non-singular matrix and therefore, $(AB)^{-1}$ exists and is given by $(AB)^{-1} = \frac{1}{|AB|} adj(AB) = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}61&-87\\-47&67\end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{r}}-61&87\\47&-67\end{array}} \right]$
Further, $|A| = 15 – 14 = 1 \neq 0$ and $|B| = 54 – 56 = -2 \neq 0$. So, A and B are both non-singular matrices and therefore, $A^{-1}$ and $B^{-1}$ both exist and are given by:
$A^{-1} = \left[ {\begin{array}{*{20}{r}}5&-7\\-2&3\end{array}} \right]$, $B^{-1} = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}9&-8\\-7&6\end{array}} \right]$
$\therefore B^{-1}A^{-1} = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}9&-8\\-7&6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}5&-7\\-2&3\end{array}} \right] = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}61&-87\\-47&67\end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{r}}-61&87\\47&-67\end{array}} \right]$
Hence, $(AB)^{-1} = B^{-1}A^{-1}$.
6. If $A = \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right]$, show that $A^2 – 5A + 7I = O$. Hence, find $A^{-1}$.
Solution:
$A = \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right]$
L.H.S. $= A^2 – 5A + 7I$
$= \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right] – 5\left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right] + 7\left[ {\begin{array}{*{20}{r}}1&0\\0&1\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}8&5\\-5&3\end{array}} \right] – \left[ {\begin{array}{*{20}{r}}15&5\\-5&10\end{array}} \right] + \left[ {\begin{array}{*{20}{r}}7&0\\0&7\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}8-15+7&5-5+0\\-5+5+0&3-10+7\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right] = O$
$\Rightarrow A^2 – 5A + 7I = O$
Hence, proved.
Now, multiplying by $A^{-1}$ on both sides, we get
$(A^{-1}A)A – 5AA^{-1} – 7IA^{-1} = O \Rightarrow IA – 5I + 7A^{-1} = O$
$\Rightarrow A – 5I + 7A^{-1} = O \Rightarrow 7A^{-1} = 5I – A$
$\Rightarrow 7A^{-1} = 5\left[ {\begin{array}{*{20}{r}}1&0\\0&1\end{array}} \right] – \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right] \Rightarrow 7A^{-1} = \left[ {\begin{array}{*{20}{r}}2&-1\\1&3\end{array}} \right]$
$\Rightarrow A^{-1} = \frac{1}{7}\left[ {\begin{array}{*{20}{r}}2&-1\\1&3\end{array}} \right]$
7. For the matrix $A = \left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right]$, find the numbers a and b such that $A^2 + aA + bI = O$.
Solution:
We are given that, $A = \left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right]$
Now, $A^2 + aA + bI = O$
$\Rightarrow \left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right] + a\left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right] + b\left[ {\begin{array}{*{20}{r}}1&0\\0&1\end{array}} \right] = O$
$\Rightarrow \left[ {\begin{array}{*{20}{r}}9+2&6+2\\3+1&2+1\end{array}} \right] + \left[ {\begin{array}{*{20}{r}}3a&2a\\a&a\end{array}} \right] + \left[ {\begin{array}{*{20}{r}}b&0\\0&b\end{array}} \right] = O$
$\Rightarrow \left[ {\begin{array}{*{20}{r}}11+3a+b&8+2a\\4+a&3+a+b\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right]$
$\Rightarrow 4 + a = 0 \Rightarrow a = -4$
Also, $3 + a + b = 0 \Rightarrow b = -3 + 4 \Rightarrow b = 1$
Hence, $a = -4, b = 1$
8. For the matrix $A = \left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right]$, show that $A^3 – 6A^2 + 5A + 11I = O$. Hence, find $A^{-1}$.
Solution:
We have $A = \left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right]$
$\therefore A^2 = AA = \left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right]$
and $A^3 = A^2A = \left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}8&7&1\\-23&27&-69\\32&-13&58\end{array}} \right]$
Now, L.H.S. $= A^3 – 6A^2 + 5A + 11I$
$= \left[ {\begin{array}{*{20}{r}}8&7&1\\-23&27&-69\\32&-13&58\end{array}} \right] – 6\left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right] + 5\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] + 11\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}8-24+5+11&7-12+5+0&1-6+5+0\\-23+18+5+0&27-48+10+11&-69+84-15+0\\32-42+10+0&-13+18-5+0&58-84+15+11\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = O \Rightarrow A^3 – 6A^2 + 5A + 11I = O$
Hence, proved.
Now, $A^3 – 6A^2 + 5A + 11I = O$
$\Rightarrow 11I = -A^3 + 6A^2 – 5A$
Multiplying (i) by $A^{-1}$, we get
$11A^{-1}I = -A^{-1}A^3 + 6A^{-1}A^2 – 5A^{-1}A \Rightarrow 11A^{-1} = -A^2 + 6A – 5I$
$\Rightarrow A^{-1} = -\frac{1}{11}A^2 + \frac{6}{11}A – \frac{5}{11}I$
$= -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right] + \frac{6}{11}\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] – \frac{5}{11}\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-4+6-5&-2+6+0&-1+6+0\\3+6+0&-8+12-5&14-18+0\\-7+12+0&3-6+0&-14+18-5\end{array}} \right]$
$= \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-3&4&5\\9&-1&-4\\5&-3&-1\end{array}} \right]$
9. If $A = \left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right]$, verify that $A^3 – 6A^2 + 9A – 4I = O$ and hence, find $A^{-1}$.
Solution:
We have $A = \left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right]$
$\therefore A^2 = AA = \left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right]$
and $A^3 = A^2A = \left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}22&-21&21\\-21&22&-21\\21&-21&22\end{array}} \right]$
Now, $A^3 – 6A^2 + 9A – 4I$
$= \left[ {\begin{array}{*{20}{r}}22&-21&21\\-21&22&-21\\21&-21&22\end{array}} \right] – 6\left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right] + 9\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] – 4\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}22-36+18-4&-21+30-9+0&21-30+9+0\\-21+30-9-0&22-36+18-4&-21+30-9+0\\21-30+9+0&-21+30-9+0&22-36+18-4\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = O$
Hence, $A^3 – 6A^2 + 9A – 4I = O$
Now, $A^3 – 6A^2 + 9A – 4I = O \Rightarrow 4I = A^3 – 6A^2 + 9A$
Multiplying both sides by $A^{-1}$, we get
$4A^{-1}I = A^{-1}A^3 – 6A^{-1}A^2 + 9A^{-1}A \Rightarrow 4A^{-1} = A^2 – 6A + 9I$
$A^{-1} = \frac{1}{4}A^2 – \frac{6}{4}A + \frac{9}{4}I$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right] – \frac{6}{4}\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] + \frac{9}{4}\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{r}}6-12+9&-5+6+0&5-6+0\\-5+6+0&6-12+9&-5+6+0\\5-6+0&-5+6+0&6-12+9\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}3&1&-1\\1&3&-1\\-1&-1&3\end{array}} \right]$
10. Let A be a non-singular square matrix of order $3 \times 3$. Then, $|adj A|$ is equal to:
Solution:
(B) $|A|^2$
For any n × n matrix A, $\det(adj A) = |A|^{n-1}$ (It holds for singular and non-singular matrices.)
11. If A is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to:
Solution:
(B) $\frac{1}{\det(A)}$
When A is an invertible matrix of order 2, $AA^{-1} = I_2 = A^{-1}A$, where $I_2$ is the identity matrix of order 2.
$\Rightarrow \det(AA^{-1}) = \det I \Rightarrow \det A \cdot \det(A^{-1}) = 1$
$\Rightarrow \det(A^{-1}) = \frac{1}{\det A}, \det A \neq 0$