Question

If

A^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]

and

B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]

, find

(A B)^{- 1}

.

Accepted Answer

We know that

$(A B)^{- 1} = B^{- 1} A^{- 1}$

We are given $A^{- 1}$ , so calculating $B^{- 1}$ .

Calculating $B^{- 1}$

We know that

$B^{- 1} = \frac{1}{|B|} adj (B)$

exists if $|B| \neq 0$

$|B| = |\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}|$

$= 1 (3 - 0) - 2 (- 1 - 0) - 2 (2 - 0) = 1 (3) - 2 (- 1) - 2 (2)$

$= 3 + 2 - 4 = 1$

Since $|B| \neq 0$ .

Thus, $B^{- 1}$ exists.

Calculating $adj B$

Now, $adj B = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}]$ where $A_{i j}$ denotes the cofactor of element $a_{i j}$ in $B$ .

$B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$

The Minor of an element:

$M_{11} = |\begin{matrix} 3 & 0 \\ - 2 & 1 \end{matrix}| = 3 (1) - (- 2) 0 = 3$

$M_{12} = |\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}| = - 1 (1) - 0 (0) = - 1$

$M_{13} = |\begin{matrix} - 1 & 3 \\ 0 & - 2 \end{matrix}| = (- 1) (- 2) - 0 (3) = 2$

$M_{21} = |\begin{matrix} 2 & - 2 \\ - 2 & 1 \end{matrix}| = 2 (1) - (- 2) (- 2) = - 2$

$M_{22} = |\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}| = 1 (1) - 0 (- 2) = 1$

$M_{23} = |\begin{matrix} 1 & 2 \\ 0 & - 2 \end{matrix}| = 1 (- 2) - 0 (2) = - 2$

$M_{31} = |\begin{matrix} 2 & - 2 \\ 3 & 0 \end{matrix}| = 2 (0) - 3 (- 2) = 6$

$M_{32} = |\begin{matrix} 1 & - 2 \\ - 1 & 0 \end{matrix}| = 1 (0) - (- 1) (- 2) = - 2$

$M_{33} = |\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}| = 1 (3) - (- 1) 2 = 5$

Now,
The Cofactors of all elements are:

$A_{11} = {(- 1)}^{1 + 1} M_{11} = {(- 1)}^{2} \cdot 3 = 3$

$A_{12} = {(- 1)}^{1 + 2} M_{12} = {(- 1)}^{3} (- 1) = 1$

$A_{13} = {(- 1)}^{1 + 3} M_{13} = {(- 1)}^{4} 2 = 2$

$A_{21} = {(- 1)}^{2 + 1} M_{21} = {(- 1)}^{3} (- 2) = 2$

$A_{22} = {(- 1)}^{2 + 2} M_{22} = {(- 1)}^{4} \cdot 1 = 1$

$A_{23} = {(- 1)}^{2 + 3} M_{23} = {(- 1)}^{5} (- 2) = 2$

$A_{31} = {(- 1)}^{3 + 1} M_{31} = (- 1)^{4} \cdot 6 = 6$

$A_{32} = {(- 1)}^{3 + 2} M_{32} = {(- 1)}^{5} (- 2) = 2$

$A_{33} = {(- 1)}^{3 + 3} M_{33} = (- 1)^{6} \cdot 5 = 5$

Thus, $adj (B) = [\begin{array}{l} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}]$ .

Now,

$B^{- 1} = \frac{1}{|B|} adj (B)$

Putting values

$= \frac{1}{1} [\begin{array}{l} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}] = [\begin{array}{l} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}]$

Also,

$(A B)^{- 1}$

$= B^{- 1} A^{- 1}$

$= [\begin{array}{l} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}] [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$

$= [\begin{array}{l} 3 (3) + 2 (- 15) + 6 (5) & 3 (- 1) + 2 (6) + 6 (- 2) & 3 (1) + 2 (- 5) + 6 (2) \\ 1 (3) + 1 (- 15) + 2 (5) & 1 (- 1) + 1 (6) + 2 (- 2) & 1 (1) + 1 (- 5) + 2 (2) \\ 2 (3) + 2 (- 15) + 5 (5) & 2 (- 1) + 2 (6) + 5 (- 2) & 2 (1) + 2 (- 5) + 5 (2) \end{array}]$

$= [\begin{matrix} 9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & - 2 + 12 - 10 & 2 - 10 + 10 \end{matrix}]$

$= [\begin{matrix} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix}]$

Mizoram Board Solutions for Chapter: Determinants, Exercise 7: Miscellaneous Exercises on Chapter 4

Mizoram Board Mathematics Solutions for Exercise - Mizoram Board Solutions for Chapter: Determinants, Exercise 7: Miscellaneous Exercises on Chapter 4

Questions from Mizoram Board Solutions for Chapter: Determinants, Exercise 7: Miscellaneous Exercises on Chapter 4 with Hints & Solutions

Learn and Prepare for any exam you want !