In the previous section, we saw adjoint and inverse of a matrix. In this section, we will see some solved examples.
Solved example 20.20
If $A~=~\left [\begin{array}{r}
1 &{ 3 } &{ 3 } \\
1 &{ 4 } &{ 3 } \\
1 &{ 3 } &{ 4 } \\
\end{array}\right ]$, then verify that A(adj A) = |A| I. Also find A−1.
Solution:
Part (i):
1. First we will find the Minors and Cofactors of A.
2. So the Cofactor matrix of A is: $\left [\begin{array}{r}
7 &{ -1 } &{ -1 } \\
-3 &{ 1 } &{ 0 } \\
-3 &{ 0 } &{ 1 } \\
\end{array}\right ]$
3. Transpose of the Cofactor matrix will give the adjoint matrix. So we get:
adj A = $\left [\begin{array}{r}
7 &{ -3 } &{ -3 } \\
-1 &{ 1 } &{ 0 } \\
-1 &{ 0 } &{ 1 } \\
\end{array}\right ]$
4. Use matrix multiplication to find A(adj A). We get:
$\left [\begin{array}{r}
1 &{ 3 } &{ 3 } \\
1 &{ 4 } &{ 3 } \\
1 &{ 3 } &{ 4 } \\
\end{array}\right ] \left [\begin{array}{r}
7 &{ -3 } &{ -3 } \\
-1 &{ 1 } &{ 0 } \\
-1 &{ 0 } &{ 1 } \\
\end{array}\right ] ~=~\left [\begin{array}{r}
1 &{ 0 } &{ 0 } \\
0 &{ 1 } &{ 0 } \\
0 &{ 0 } &{ 1 } \\
\end{array}\right ]$
5. Let us calculate |A|:
|A| = 1(16 − 9) − 3(4 − 3) + 3(3 − 4)
= 7 − 3 − 3 = 1
6. Now we can calculate |A|I:
$(1) \left [\begin{array}{r}
1 &{ 0 } &{ 0 } \\
0 &{ 1 } &{ 0 } \\
0 &{ 0 } &{ 1 } \\
\end{array}\right ] ~=~\left [\begin{array}{r}
1 &{ 0 } &{ 0 } \\
0 &{ 1 } &{ 0 } \\
0 &{ 0 } &{ 1 } \\
\end{array}\right ]$
7. From (4) and (6), we see that: A(adj A) = |A| I
Part (ii):
We have already calculated |A| and (adj A) in part (i). So we can calculate A−1 as follows:
$A^{-1} = \frac{1}{|A|}(\text{adj A}) = \frac{1}{1} \left [\begin{array}{r}
7 &{ -3 } &{ -3 } \\
-1 &{ 1 } &{ 0 } \\
-1 &{ 0 } &{ 1 } \\
\end{array}\right ]= \left [\begin{array}{r}
7 &{ -3 } &{ -3 } \\
-1 &{ 1 } &{ 0 } \\
-1 &{ 0 } &{ 1 } \\
\end{array}\right ]$
Solved example 20.21
If $A~=~\left [\begin{array}{r}
2 &{ 3 } \\
1 &{ -4 } \\
\end{array}\right ]$ and $B~=~\left [\begin{array}{r}
1 &{ -2 } \\
-1 &{ 3 } \\
\end{array}\right ]$, then verify that (AB)−1 = B−1A−1.
Solution:
1. First we will test whether A and B are invertible.
(i) |A| = −8 − 3 = −11
|A| ≠ 0. So A is invertible.
(ii) |B| = 3-2 = 1
|B| ≠ 0. So B is invertible.
2. Next we will find A−1.
• adj A = $\left [\begin{array}{r}
-4 &{ -3 } \\
-1 &{ 2 } \\
\end{array}\right ]$
(Recall the shortcut method explained in fig.20.5 of the previous section)
• So we get:
$A^{-1} = \frac{1}{|A|}(\text{adj A}) = -{\frac{1}{11}} \left [\begin{array}{r}
-4 &{ -3 } \\
-1 &{ 2 } \\
\end{array}\right ]$
3. Next we will find B−1.
• adj B = $\left [\begin{array}{r}
3 &{ 2 } \\
1 &{ 1 } \\
\end{array}\right ]$
(Recall the shortcut method explained in fig.20.5 of the previous section)
• So we get:
$B^{-1} = \frac{1}{|B|}(\text{adj B}) = \frac{1}{1} \left [\begin{array}{r}
3 &{ 2 } \\
1 &{ 1 } \\
\end{array}\right ]= \left [\begin{array}{r}
3 &{ 2 } \\
1 &{ 1 } \\
\end{array}\right ]$
4. Use matrix multiplication to find B−1A−1. We get:
$B^{-1} A^{-1} = \left [\begin{array}{r}
3 &{ 2 } \\
1 &{ 1 } \\
\end{array}\right ]~\times~ -{\frac{1}{11}} \left [\begin{array}{r}
-4 &{ -3 } \\
-1 &{ 2 } \\
\end{array}\right ] = {\frac{1}{11}} \left [\begin{array}{r}
14 &{ 5 } \\
5 &{ 1 } \\
\end{array}\right ]$
5. Use matrix multiplication to find AB. We get:
$AB = \left [\begin{array}{r}
2 &{ 3 } \\
1 &{ -4 } \\
\end{array}\right ]~\left [\begin{array}{r}
1 &{ -2 } \\
-1 &{ 3 } \\
\end{array}\right ] = \left [\begin{array}{r}
-1 &{ 5 } \\
5 &{ -14 } \\
\end{array}\right ]$
6. Next we will find |(AB)|.
|(AB)| = 14 − 25 = -11
7. Now we can calculate (AB)−1.
• adj (AB) = $\left [\begin{array}{r}
-14 &{ -5 } \\
-5 &{ -1 } \\
\end{array}\right ]$
(Recall the shortcut method explained in fig.20.5 of the previous section)
• So we get:
$(AB)^{-1} = \frac{1}{|(AB)|}(\text{adj (AB)}) = \frac{1}{(-11)} \left [\begin{array}{r}
-14 &{ -5 } \\
-5 &{ -1 } \\
\end{array}\right ] = \frac{1}{11} \left [\begin{array}{r}
14 &{ 5 } \\
5 &{ 1 } \\
\end{array}\right ]$
8. Comparing the results in (4) and (7), we can write:
(AB)−1 = B−1A−1.
Solved example 20.22
Show that the matrix $A~=~\left [\begin{array}{r}
2 &{ 3 } \\
1 &{ 2 } \\
\end{array}\right ]$ satisfies the equation A2 − 4A + I = O, where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A−1.
Solution:
Part (i): To prove that, A2 − 4A + I = O
1. Use matrix multiplication to find A2. We get:
$A^2 = AA = \left [\begin{array}{r}
2 &{ 3 } \\
1 &{ 2 } \\
\end{array}\right ]~\left [\begin{array}{r}
2 &{ 3 } \\
1 &{ 2 } \\
\end{array}\right ] = \left [\begin{array}{r}
7 &{ 12 } \\
4 &{ 7 } \\
\end{array}\right ]$
2. Next we will find 4A.
$4A = 4 \left [\begin{array}{r}
2 &{ 3 } \\
1 &{ 2 } \\
\end{array}\right ]= \left [\begin{array}{r}
8 &{ 12 } \\
4 &{ 8 } \\
\end{array}\right ]$
3. Substituting the above matrices in the given equation, we get:
• We see that, L.H.S = R.H.S.
• So matrix A satisfies the given equation.
Part (ii): To find A−1.
1. First we need to test whether A is invertible.
• We have: |A| = (4-3) = 1.
• Since |A| ≠ 0, A is invertible.
2. To find A−1, we rearrange the equation that was proved in part (i).
◼ Remarks:
• 3 (magenta color): In this line, we post multiply the whole equation by A−1.
• 4 (magenta color): In this line, we apply the fact that, AA−1 = I.
The link below gives a few more solved examples:
In the next section, we will see the applications of determinants and matrices.
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