Question

If A = [ 2 3 0 −2 ] and B = [ −8 8 ], find matrix X such that 2AX = B.

Answer 1

GIVEN :

The matrices are $A=\left[\begin{array}{cc}2&-3\\0&-2\end{array}\right]$ and $B=\left[\begin{array}{c}-8\\8\end{array}\right]$, find matrix X such that 2AX = B.

TO FIND :

The value of the matrix X

SOLUTION :

Given matrices are

$A=\left[\begin{array}{cc}2&-3\\0&-2\end{array}\right]$ and

$B=\left[\begin{array}{c}-8\\8\end{array}\right]$

Also given that 2AX=B

We have that $X=\frac{1}{2}A^{-1}B$

First find the matrix  $A^{-1}$

$A^{-1}=\frac{adjA}{|A|}$

$adjA=\left[\begin{array}{cc}-2&-3\\0&2\end{array}\right]$

$|A|=\left|\begin{array}{cc}2&-3\\0&-2\end{array}\right|$

= -4-0

=-4

⇒ |A| = -4

Now $A^{-1}=\frac{adjA}{|A|}$

$A^{-1}=\frac{\left[\begin{array}{cc}-2&-3\\0&2\end{array}\right]}{-4}$

$A^{-1}=\left[\begin{array}{cc}\frac{1}{2}&\frac{3}{4}\\0&-\frac{1}{2}\end{array}\right]$

Substituting the matrices in $X=\frac{1}{2}A^{-1}B$ we get

$X=\frac{1}{2}\left[\begin{array}{cc}\frac{1}{2}&\frac{3}{4}\\0&-\frac{1}{2}\end{array}\right]\times \left[\begin{array}{c}-8\\8\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{cc}-4&-6\\0&-4\end{array}\right]$

$X=\left[\begin{array}{cc}-2&-3\\0&-2\end{array}\right]$

∴ the value of the matrix X is $\left[\begin{array}{cc}-2&-3\\0&-2\end{array}\right]$

Answer 2

Step-by-step explanation:

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