Question

Find the matrix of the given question

Answer 1

We're given,

$\mathrm{\left(\begin{array}{cc}2&1\\-3&4\end{array}\right)\cdot X=\left(\begin{array}{c}7\\6\end{array}\right)}$

Here a 2 × 2 matrix is multiplied with an unknown matrix X to get a 2 × 1 matrix.

So the matrix X should also be a 2 × 1 matrix, because we know that the product of an m × n matrix and an n × p matrix in the order gives an m × p matrix.

Hence the order of X is 2 × 1.

Let,

$X=\left(\begin{array}{c}a\\b\end{array}\right).$

So,

$\begin{aligned}&\left(\begin{array}{cc}2&1\\-3&4\end{array}\right)\cdot\left(\begin{array}{c}a\\b\end{array}\right)=\left(\begin{array}{c}7\\6\end{array}\right)\\\\\implies\ \ &\left(\begin{array}{c}2a+b\\-3a+4b\end{array}\right)=\left(\begin{array}{c}7\\6\end{array}\right)\end{aligned}$

Here we get two equations.

$\begin{aligned}2a+b&=7\\-3a+4b&=6\end{aligned}$

Let's solve it by matrix method!

We have the given matrix to solve,

$\left(\begin{array}{cc|c}2&1&7\\-3&4&6\end{array}\right)$

Now,

$\left(\begin{array}{cc|c}2&1&7\\-3&4&6\end{array}\right)\xrightarrow{R_1+R_2\to R_1}\left(\begin{array}{cc|c}-1&5&13\\-3&4&6\end{array}\right)\xrightarrow{-R_1\to R_1}\left(\begin{array}{cc|c}1&-5&-13\\-3&4&6\end{array}\right)\\\\\\\xrightarrow{3R_1+R_2\to R_2}\left(\begin{array}{cc|c}1&-5&-13\\0&-11&-33\end{array}\right)\xrightarrow{-\frac{1}{11}R_2\to R_2}\left(\begin{array}{cc|c}1&-5&-13\\0&1&3\end{array}\right)\\\\\\\xrightarrow{R_1+5R_2\to R_1}\left(\begin{array}{cc|c}1&0&2\\0&1&3\end{array}\right)$

This implies,

$\begin{aligned}a&=2\\b&=3\end{aligned}$

Hence,

$\mathrm{X=\left(\begin{array}{c}2\\3\end{array}\right)}$