20.13 - Solved Examples

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.

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The link below gives a few more solved examples:

Exercise 20.5

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In the next section, we will see the applications of determinants and matrices.

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