Question

Problem 3: Find the matrix "X" that makes the equation true: 11 ЗХ. [? -13 15 -19 1 2

Answer 1

The matrix is given as :

$3\cdot x\text{ -}\begin{bmatrix}{11} & {-6} & {} \\ {2} & {1} & \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-13} & {15} & {} \\ {-19} & {2} & {} \\ {} & {} & {}\end{bmatrix}$

You should find matrix x that multiplies by 3 and subtracts the second matrix to get the answer matrix

Add the second matrix and the answer matrix and equate to 3x as;

$3x\text{ =}\begin{bmatrix}{11} & {-6} & {} \\ {2} & {1} & {} \\ {} & {} & \end{bmatrix}+\begin{bmatrix}{-13} & {15} & \\ {-19} & {2} & {} \\ {} & {} & {}\end{bmatrix}$

This will give;

$3x\text{ = }\begin{bmatrix}{11-13} & {-6+15} & {} \\ {2-19} & {1+2} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-2} & {9} & {} \\ {-17} & {3} & {} \\ {} & {} & {}\end{bmatrix}$

Divide every value in the answer matrix by 3 to get the x matrix as;

$x=\begin{bmatrix}{-\frac{2}{3}} & {\frac{9}{3}} & {} \\ {-\frac{17}{3}} & {\frac{3}{3}} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-\frac{2}{3}} & {3} & \\ {-\frac{17}{3}} & {1} & {} \\ {} & {} & {}\end{bmatrix}$

Answer is :

$x=\begin{bmatrix}{-\frac{2}{3}} & {3} & {} \\ {-\frac{17}{3}} & {1} & {} \\ {} & {} & {}\end{bmatrix}$