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Found in: Page 39

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# There exists a${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix such that ${\mathbit{A}}\left[\begin{array}{c}1\\ 2\end{array}\right]{\mathbf{=}}\left[\begin{array}{c}3\\ 4\end{array}\right]$

True, there exist a$2×2$ matrix A such that role="math" localid="1659642050851" $A\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$

See the step by step solution

## Step 1:Taking the matrix

Suppose the matrix A is.

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Now according to the given equation,

$A\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$

## Step 2: Justification of answer

Put the value of A matrix in the above equation we get.

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$

After solving the equation we get

$\left[\begin{array}{c}a+2b\\ c+2d\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$

Now on equating the value of matrix we get

$a+2b=3\phantom{\rule{0ex}{0ex}}c+2d=4\phantom{\rule{0ex}{0ex}}$

On solving the equations we get the value of variables

$a=3-2b\phantom{\rule{0ex}{0ex}}c=4-2d\phantom{\rule{0ex}{0ex}}$

Where, are free variables which can have any value. For instance let the value of $b,d=1$

It implies $a=1,c=2$

Thus, we get the matrix

$A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$

Hence, it is truethat there exist a $2×2$matrix A such that $A\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$