floymdiT

2021-02-16

If $3\left[\begin{array}{cc}{x}_{1}& {x}_{2}\\ {x}_{3}& {x}_{4}\end{array}\right]=\left[\begin{array}{cc}{x}_{1}& 2\\ -1& 4{x}_{4}\end{array}\right]+\left[\begin{array}{cc}4& {x}_{1}+{x}_{2}\\ {x}_{3}+{x}_{4}& 3\end{array}\right]$
1. ${x}_{1}=-2,{x}_{2}=2,{x}_{3}=-2,{x}_{4}=-3$
2. ${x}_{1}=2,{x}_{2}=-2,{x}_{3}=-2,{x}_{4}=-3$
3. ${x}_{1}=2,{x}_{2}=2,{x}_{3}=2,{x}_{4}=-3$
4. ${x}_{1}=2,{x}_{2}=2,{x}_{3}=-2,{x}_{4}=3$
5, ${x}_{1}=2,{x}_{2}=2,{x}_{3}=-2,{x}_{4}=-3$

Latisha Oneil

Step 1
We will perform the addition and scalar multiplication of matrices on both the sides.
Then we will compare corresponding elements in both matrices and will solve the equations.
Step 2
$3\left[\begin{array}{cc}{x}_{1}& {x}_{2}\\ {x}_{3}& {x}_{4}\end{array}\right]=\left[\begin{array}{cc}{x}_{1}& 2\\ -1& 4{x}_{4}\end{array}\right]+\left[\begin{array}{cc}3& {x}_{1}+{x}_{2}\\ {x}_{3}+{x}_{4}& 3\end{array}\right]$
$\therefore \left[\begin{array}{cc}3{x}_{1}& 3{x}_{2}\\ 3{x}_{3}& 3{x}_{4}\end{array}\right]=\left[\begin{array}{cc}{x}_{1}+4& 2+{x}_{1}+{x}_{2}\\ {x}_{3}+{x}_{4}-1& 4{x}_{4}+3\end{array}\right]$
$\therefore 3{x}_{1}={x}_{1}+4$
$2{x}_{1}=4$
${x}_{1}=2$
$\therefore 3{x}_{4}=4{x}_{4}+3$
${x}_{4}+3=0$
${x}_{4}=-3$
$\therefore 3{x}_{2}=2+{x}_{1}+{x}_{2}$
$2{x}_{2}=2+{x}_{1}$
$2{x}_{2}=2+2$
${x}_{2}=2$
$\therefore 3{x}_{3}={x}_{3}+{x}_{4}-1$
$2{x}_{3}={x}_{4}-1$
$=-3-1$
${x}_{3}=-2$
$\therefore {x}_{1}=2,{x}_{2}=2,{x}_{3}=-2,{x}_{4}=-3$
option 5

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