Question

please answer this question please​

Answer 1

Answer:

$X=\begin{bmatrix}3&4\\-1&7\end{bmatrix}$

Step-by-step explanation:

Given,

$A=\begin{bmatrix}0&-1\\1&2\end{bmatrix}$

$B=\begin{bmatrix}1&2\\-1&1\end{bmatrix}$

X - 3B = 2A

To Find :-

Matrix 'X'

How To Do :-

As they given the condition that 'X - 3B = 2A' we need to substitute the value of the matrices 'B' and 'A' in that equation and we need to simplify them by transposing and we need to obtain the value of matrix 'X'.

Formula Required :-

$1)y\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ay&by\\cy&dy\end{bmatrix}$

$2)\begin{bmatrix}a & b \\c & d \end{bmatrix}+\begin{bmatrix}p & q \\r & s \end{bmatrix}=\begin{bmatrix}a+p & b+q \\c+r &d+ s \end{bmatrix}$

Solution :-

X - 3B = 2A

Substituting the matrices in the above equation :-

$X-3\begin{bmatrix}1&2\\-1&1\end{bmatrix}=2\begin{bmatrix}0&-1\\1&2\end{bmatrix}$

$X-\begin{bmatrix}3\times1&3\times2\\3\times-1&3\times1\end{bmatrix}=\begin{bmatrix}2\times0&2\times-1\\2\times1&2\times2\end{bmatrix}$

$X-\begin{bmatrix}3&6\\-3&3\end{bmatrix}=\begin{bmatrix}0&-2\\2&4\end{bmatrix}$

$X=\begin{bmatrix}0&-2\\2&4\end{bmatrix}+\begin{bmatrix}3&6\\-3&3\end{bmatrix}$

$X=\begin{bmatrix}0+3&-2+6\\2+(-3)&4+3\end{bmatrix}$

$X=\begin{bmatrix}3&4\\-1&7\end{bmatrix}$

$\therefore X=\begin{bmatrix}3&4\\-1&7\end{bmatrix}$

Answer 2

Answer :-

X =  $\left[\begin{array}{ccc}3&4\\-1&7\end{array}\right]$

Step-by-step explanation :-

Given :

If A = $\left[\begin{array}{ccc}0&-1\\1&2\end{array}\right]$ and B = $\left[\begin{array}{ccc}1&2\\-1&1\end{array}\right]$.

To find :

The matrix X if X - 3B = 2A.

Solution :

3B :-

3 $\times$  $\left[\begin{array}{ccc}1&2\\-1&1\end{array}\right]$ = $\left[\begin{array}{ccc}3&6\\-3&3\end{array}\right]$

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2A :-

2 $\times$  $\left[\begin{array}{ccc}0&-1\\1&2\end{array}\right]$ = $\left[\begin{array}{ccc}0&-2\\2&4\end{array}\right]$

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X - 3B = 2A :-

X -  $\left[\begin{array}{ccc}3&6\\-3&3\end{array}\right]$ =  $\left[\begin{array}{ccc}0&-2\\2&4\end{array}\right]$

X = $\left[\begin{array}{ccc}0&-2\\2&4\end{array}\right]$ +  $\left[\begin{array}{ccc}3&6\\-3&3\end{array}\right]$

X = $\left[\begin{array}{ccc}(0+3)&(-2+6)\\(2+(-3))&(4+3)\end{array}\right]$

∴ X = $\left[\begin{array}{ccc}3&4\\-1&7\end{array}\right]$

Verification :

X - 3B = 2A

$\left[\begin{array}{ccc}3&4\\-1&7\end{array}\right]$ -  $\left[\begin{array}{ccc}3&6\\-3&3\end{array}\right]$ =  $\left[\begin{array}{ccc}0&-2\\2&4\end{array}\right]$

$\left[\begin{array}{ccc}(3-3)&(4-6)\\(-1)-(-3)&(7-4)\\\end{array}\right]$

 $\left[\begin{array}{ccc}0&-2\\2&4\end{array}\right]$ =   $\left[\begin{array}{ccc}0&-2\\2&4\end{array}\right]$

∴ L.H.S = R.H.S