Name: The Applications of Matrices
Uploaded: 2019-07-19
Duration: 7 min 47 s
Description: Let's learn the applications of matrices, such as solving homogenous and non-homogenous linear equations, the Gauss-Jordan method, and the reduction method.

Question

If

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

find

A B and {(A B)}^{- 1}

. Verify that

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

.

Accepted Answer

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

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

$A B = [\begin{matrix} (2 \times 1) + (3 \times 3) & (2 \times 0) + (3 \times 1) \\ (1 \times 1) + (2 \times 3) & (1 \times 0) + (2 \times 1) \end{matrix}]$

$A B = [\begin{matrix} 11 & 3 \\ 7 & 2 \end{matrix}]$

Now, ${(A B)}^{- 1} = \frac{1}{|A B|} [\begin{matrix} 2 & - 3 \\ - 7 & 11 \end{matrix}]$ ...........(i)

$|A B| = (11 \times 2) - (3 \times 7)$

$|A B| = 22 - 21$

$|A B| = 1$

From (i), ${(A B)}^{- 1} = \frac{1}{1} [\begin{matrix} 2 & - 3 \\ - 7 & 11 \end{matrix}]$

Therefore, ${(A B)}^{- 1} = [\begin{matrix} 2 & - 3 \\ - 7 & 11 \end{matrix}]$

Now, $A^{- 1} = \frac{1}{|A|} [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}]$ ........(ii)

$|A| = (2 \times 2) - (3 \times 1)$

$|A| = 1$

From (ii), $A^{- 1} = \frac{1}{1} [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}]$

$∴ A^{- 1} = [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}]$

Now, $B^{- 1} = \frac{1}{|B|} [\begin{matrix} 1 & - 0 \\ - 3 & 1 \end{matrix}]$ ........(iiI)

$|B| = (1 \times 1) - (- 3 \times - 0)$

$|B| = 1$

From (iii), $B^{- 1} = \frac{1}{1} [\begin{matrix} 1 & - 0 \\ - 3 & 1 \end{matrix}]$

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

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

$B^{- 1} A^{- 1} = [\begin{matrix} (1 \times 2) + (0 \times - 1) & (1 \times - 3) + (0 \times 2) \\ (- 3 \times 2) + (1 \times - 1) & (- 3 \times - 3) + (1 \times 2) \end{matrix}]$

$B^{- 1} A^{- 1} = [\begin{matrix} 2 & - 3 \\ - 7 & 11 \end{matrix}]$

Hence, proved that ${(A B)}^{- 1} = B^{- 1} A^{- 1}$

If $A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix}]$ find $A B and {(A B)}^{- 1}$ . Verify that ${(A B)}^{- 1} = B^{- 1} A^{- 1}$ .

Important Questions on Matrices

Learn and Prepare for any exam you want !

If A=2312, B=1031 find AB and AB-1. Verify that AB-1=B-1A-1.

Important Questions on Matrices

Learn and Prepare for any exam you want !

If $A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix}]$ find $A B and {(A B)}^{- 1}$ . Verify that ${(A B)}^{- 1} = B^{- 1} A^{- 1}$ .