# 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.

Let It can be shown that $4u-3v-w=0$. Use this fact (and no row operations) to find a solution to the system $4u-3v-w=0$ , where
$A=\left[\begin{array}{cc}2& -4\\ 5& 17\\ -1& -13\end{array}\right],x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right],b=\left[\begin{array}{c}4\\ 1\\ 3\end{array}\right]$

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Obiajulu
Step 1
Every matrix has 2 attributes, rows and columns, if a matrix A is represented as ${A}_{mxn}$, then m represents the number of rows and n represents the number of columns present in the matrix. In order to add two matrices, A and B, the number of columns of matrix A should be equal to the number of rows of matrix B.
Two or more matrices can be used to solve linear equations by equating, using row or column operations in order to reduce the matrix and there are a lot of other ways as well. Identity matrix is a special kind of matrix with only 1 on its diagonal elements.
Step 2
Here,

We are also given:
$4u-3v-w=0$
or
$4u-w=3v$
$⇒\frac{4}{3}u-\frac{1}{3}w=v$
According to vector equation is: $\left[\begin{array}{cc}2& -4\\ 5& 17\\ -1& -13\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}4\\ 1\\ 3\end{array}\right]$
Hence, according to the vector equation we get
${x}_{1}=\frac{4}{3}$
${x}_{2}=\frac{1}{3}$