# If 3begin{bmatrix}x_1 & x_2 x_3 & x_4 end{bmatrix}=begin{bmatrix}x_1 & 2 -1 & 4x_4 end{bmatrix}+begin{bmatrix}4 & x_1+x_2 x_3+x_4 & 3 end{bmatrix} 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

Question
Matrices
If $$3\begin{bmatrix}x_1 & x_2 \\x_3 & x_4 \end{bmatrix}=\begin{bmatrix}x_1 & 2 \\-1 & 4x_4 \end{bmatrix}+\begin{bmatrix}4 & x_1+x_2 \\x_3+x_4 & 3 \end{bmatrix}$$
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, $$x_1=2 , x_2=2 , x_3=-2 , x_4=-3$$

2021-02-17
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
$$3\begin{bmatrix}x_1 & x_2 \\x_3 & x_4 \end{bmatrix}=\begin{bmatrix}x_1 & 2 \\-1 & 4x_4 \end{bmatrix}+\begin{bmatrix}3 & x_1+x_2 \\x_3+x_4 & 3 \end{bmatrix}$$
$$\therefore \begin{bmatrix}3x_1 & 3x_2 \\3x_3 & 3x_4 \end{bmatrix}=\begin{bmatrix}x_1+4 & 2+x_1+x_2 \\x_3+x_4-1 & 4x_4+3 \end{bmatrix}$$
$$\therefore 3x_1=x_1+4$$
$$2x_1=4$$
$$x_1=2$$
$$\therefore 3x_4=4x_4+3$$
$$x_4+3=0$$
$$x_4=-3$$
$$\therefore 3x_2=2+x_1+x_2$$
$$2x_2=2+x_1$$
$$2x_2=2+2$$
$$x_2=2$$
$$\therefore 3x_3=x_3+x_4-1$$
$$2x_3=x_4-1$$
$$=-3-1$$
$$x_3=-2$$
$$\therefore x_1=2,x_2=2,x_3=-2,x_4=-3$$
option 5

### Relevant Questions

The row echelon form of a system of linear equations is given.
(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or $$x_1,x_2,x_3, x_4$$
(b) Determine whether the system is consistent. If it is consistent, give the solution.
$$\begin{matrix}1 & 0 & 3 & 0 &1 \\ 0 & 1 & 4 & 3&2\\0&0&1&2&3\\0&0&0&0&0 \end{matrix}$$
The row echelon form of a system of linear equations is given.
(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or $$x_1,x_2,x_3, x_4$$
(b) Determine whether the system is consistent. If it is consistent, give the solution.
$$\begin{matrix}1 & 0 & 2 & -1 \\ 0 & 1 & -4 & -2\\0&0&0&0&0 \end{matrix}$$
Let $$u=\begin{bmatrix}2 \\ 5 \\ -1 \end{bmatrix} , v=\begin{bmatrix}4 \\ 1 \\ 3 \end{bmatrix} \text{ and } w=\begin{bmatrix}-4 \\ 17 \\ -13 \end{bmatrix}$$ It can be shown that 4u-3v-w=0. Use this fact (and no row operations) to find a solution to the system Ax=b , where
$$A=\begin{bmatrix}2 & -4 \\5 & 17\\-1&-13 \end{bmatrix} , x=\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} , b=\begin{bmatrix}4 \\ 1 \\ 3 \end{bmatrix}$$
Compute the indicated matrices, if possible .
A^2B
let $$A=\begin{bmatrix}1 & 2 \\3 & 5 \end{bmatrix} \text{ and } B=\begin{bmatrix}2 & 0 & -1 \\3 & -3 & 4 \end{bmatrix}$$
Perform the indicated matrix operations B - A given that A, B, and C are defined as follows. If an operation is not defined, state the reason.
$$A=\begin{bmatrix}4 & 0 \\-3 & 5 \\ 0 & 1 \end{bmatrix} B=\begin{bmatrix}5 & 1 \\-2 & -2 \end{bmatrix} C=\begin{bmatrix}1 & -1 \\-1 & 1 \end{bmatrix}$$
compute the indicated matrices (if possible). D+BC
Let $$A=\begin{bmatrix}3 & 0 \\ -1 & 5 \end{bmatrix} , B=\begin{bmatrix}4 & -2 & 1 \\ 0 & 2 &3 \end{bmatrix} , C=\begin{bmatrix}1 & 2 \\ 3 & 4 \\ 5 &6 \end{bmatrix} , D=\begin{bmatrix}0 & -3 \\ -2 & 1 \end{bmatrix} , E=\begin{bmatrix}4 & 2 \end{bmatrix} , F=\begin{bmatrix}-1 \\ 2 \end{bmatrix}$$
Use the graphing calculator to solve if possible
A=\begin{bmatrix}1 & 0&5 \\1 & -5&7\\0&3&-4 \end{bmatrix}\\ B=\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}\\ C=\begin{bmatrix}5 & 2&3 \\2& -1&0 \end{bmatrix}\\ D=\begin{bmatrix}5 \\-3\\4 \end{bmatrix}
Find the value in row 2 column 3 of AB-3B
compute the indicated matrices (if possible). B - C
Let
$$A=\begin{bmatrix}3 & 0 \\-1 & 5 \end{bmatrix} , B=\begin{bmatrix}4 & -2&1 \\0 & 2&3 \end{bmatrix} , C=\begin{bmatrix}1 & 2 \\3 & 4\\5&6 \end{bmatrix}, D=\begin{bmatrix}0 & -3 \\-2 & 1 \end{bmatrix},E=\begin{bmatrix}4 & 2 \end{bmatrix},F=\begin{bmatrix}-1 \\2 \end{bmatrix}$$
Solve for x and y $$\begin{bmatrix}x & 2y \\ 4 & 6 \end{bmatrix}=\begin{bmatrix}2 & -2 \\ 2x & -6y \end{bmatrix}$$
$$3\begin{bmatrix}x & y \\ y & x \end{bmatrix}=\begin{bmatrix}6 & -9 \\ -9 & 6 \end{bmatrix}$$
$$2\begin{bmatrix}x & y \\ x+y & x-y \end{bmatrix}=\begin{bmatrix}2 & -4 \\ -2 & 6 \end{bmatrix}$$
$$\begin{bmatrix}x & y \\ -y & x \end{bmatrix}-\begin{bmatrix}y & x \\ x & -y \end{bmatrix}=\begin{bmatrix}4 & -4 \\ -6 & 6 \end{bmatrix}$$
$$A=\begin{bmatrix}5 & 3 \\ -3 & -1 \\ -2 & -5 \end{bmatrix} \text{ and } B=\begin{bmatrix}0 & -2 \\ 1 & 3 \\ 4 & -3 \end{bmatrix}$$