If AB = {bmatrix} 20 & 6 \\ 7 & 3 {bmatrix}, A = {bmatrix} 4 & 1 \\ 7 & 2 {bmatrix} find B

Correct Answer is : [339−112−30]{bmatrix} 33 & 9 \\ -112 & -30 {bmatrix}[33−1129−30]

{bmatrix} 33 & 9 \\ -112 & -30 {bmatrix}

{bmatrix} 9 & 33 \\ -112 & -30 {bmatrix}

{bmatrix} 33 & 9 \\ -30 & -112 {bmatrix}

{bmatrix} 9 & 33 \\ -10 & -112 {bmatrix}

Correct Answer is : [339−112−30]{bmatrix} 33 & 9 \\ -112 & -30 {bmatrix}[33−1129−30]

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Question : 23 of 36

Marks: +1, -0

AB = \begin{bmatrix} 20 & 6 \\ 7 & 3 \end{bmatrix}, A = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix}

B

Solution:

Concept:

Given

A

and

AB

, find

B

by left-multiplying both sides by

A^{-1}

B = A^{-1} (AB)

Explanation:

First, compute

A^{-1}

for

A = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix}

\det(A) = 4 \cdot 2 - 1 \cdot 7 = 8 - 7 = 1

A^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix}

Now multiply

A^{-1}

AB = \begin{bmatrix} 20 & 6 \\ 7 & 3 \end{bmatrix}

B = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \begin{bmatrix} 20 & 6 \\ 7 & 3 \end{bmatrix} = \begin{bmatrix} 2\cdot20 + (-1)\cdot7 & 2\cdot6 + (-1)\cdot3 \\ (-7)\cdot20 + 4\cdot7 & (-7)\cdot6 + 4\cdot3 \end{bmatrix} = \begin{bmatrix} 40-7 & 12-3 \\ -140+28 & -42+12 \end{bmatrix} = \begin{bmatrix} 33 & 9 \\ -112 & -30 \end{bmatrix}

Answer:

\begin{bmatrix} 33 & 9 \\ -112 & -30 \end{bmatrix}

, which corresponds to option A.

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