Name: Finding Inverse by Elementary Transformations
Uploaded: 2019-07-19
Duration: 12 min 5 s
Description: In this video, we will learn to find the inverse of matrices using elementary transformations of matrices. We will learn how to find the inverse of a matrix,...

Question

If $A = [\begin{array}{r} 2 & 1 \\ - 1 & 3 \end{array}]$ , $B = [\begin{array}{l} 2 & 3 \\ 1 & 1 \end{array}]$ , and $C = [\begin{matrix} 1 & 0 & 2 \\ - 2 & 3 & 0 \end{matrix}]$ , verify $(A + B) C = A C + B C$

Accepted Answer

Given, $A = [\begin{array}{r} 2 & 1 \\ - 1 & 3 \end{array}], B = [\begin{array}{l} 2 & 3 \\ 1 & 1 \end{array}]$ , and $C = [\begin{matrix} 1 & 0 & 2 \\ - 2 & 3 & 0 \end{matrix}]$

To Prove : $(A + B) C = A C + B C$

Consider L.H.S.

$(A + B) C = ([\begin{array}{r} 2 & 1 \\ - 1 & 3 \end{array}] + [\begin{array}{l} 2 & 3 \\ 1 & 1 \end{array}]) [\begin{matrix} 1 & 0 & 2 \\ - 2 & 3 & 0 \end{matrix}]$

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

$= [\begin{matrix} 4 & 4 \\ 0 & 4 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ - 2 & 3 & 0 \end{matrix}]$

$= [\begin{matrix} (4 \times 1) + (4 \times (- 2)) & (4 \times 0) + (4 \times 3) & (4 \times 2) + (4 \times 0) \\ (0 \times 1) + (4 \times (- 2)) & (0 \times 0) + (4 \times 3) & (0 \times 2) + (4 \times 0) \end{matrix}]$

$= [\begin{matrix} 4 - 8 & 0 + 12 & 8 + 0 \\ 0 - 8 & 0 + 12 & 0 \end{matrix}]$

$= [\begin{matrix} - 4 & 12 & 8 \\ - 8 & 12 & 0 \end{matrix}]$ $\to (i)$

Now, consider

$A C + B C = [\begin{matrix} 2 & 1 \\ - 1 & 3 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ - 2 & 3 & 0 \end{matrix}] + [\begin{array}{l} 2 & 3 \\ 1 & 1 \end{array}] [\begin{matrix} 1 & 0 & 2 \\ - 2 & 3 & 0 \end{matrix}]$

$= [\begin{matrix} 2 - 2 & 0 + 3 & 4 \\ - 1 - 6 & 0 + 9 & - 2 + 0 \end{matrix}] + [\begin{matrix} 2 - 6 & 0 + 9 & 4 + 0 \\ 1 - 2 & 0 + 3 & 2 + 0 \end{matrix}]$

$= [\begin{matrix} 0 & 3 & 4 \\ - 7 & 9 & - 2 \end{matrix}] + [\begin{matrix} - 4 & 9 & 4 \\ - 1 & 3 & 2 \end{matrix}]$

$= [\begin{matrix} 0 - 4 & 3 + 9 & 4 + 4 \\ - 7 - 1 & 9 + 3 & - 2 + 2 \end{matrix}]$

$= [\begin{matrix} - 4 & 12 & 8 \\ - 8 & 12 & 0 \end{matrix}]$ $\to (i i)$

From $(i)$ , and $(i i)$ , we have

$(A + B) C = A C + B C$

Hence, Proved.

Odisha Board Solutions for Chapter: Matrices, Exercise 1: EXERCISES 4 (a)

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Matrices, Exercise 1: EXERCISES 4 (a)

Questions from Odisha Board Solutions for Chapter: Matrices, Exercise 1: EXERCISES 4 (a) with Hints & Solutions

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