# Equation with multiplication of a matrix by a column vector

How do I solve this matrix equation?

$$\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 4\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}5\\7\end{bmatrix}.$$

I know each line represents an equation but what should I do with the zero in the first row, second column?

• Typesetting is your friend! – David H Apr 6 '14 at 12:19

First, multiply the left hand side. You can treat this as two matrices. One is 2x3 (2 rows one column), and one is 3x1 (3 rows one column). When multiplying these matrices the end results would be a matrix whose size is 2x1.

so, multiplying row 1 by column 1, and then row 2 by column 1 gives us

\begin{bmatrix} 1\cdot x_1 + 0\cdot x_2 + 3\cdot x_3\\ 0\cdot x_1 + 1\cdot x_2 +4\cdot x_3 \end{bmatrix}

and so

$$\begin{bmatrix}1\cdot x_1 + 3\cdot x_3\\ 1\cdot x_2 +4\cdot x_3\end{bmatrix}=\begin{bmatrix} 5\\ 7\end{bmatrix}$$

See more about multiplying: Multiplying a matrix by a column vector

You can then say that for the left hand side to be equal to the right hand side you need all the elements in the matrices to be equal. Both matrices are of size 2x1, and you get two simple equations.

$$x_1 + 3\cdot x_3 = 5$$ $$x_2 + 4\cdot x_3 = 7$$

You have a system of equations with infinite solutions. I assume you need to find the general solution:

$$(x_1 =5-3x_3, x_2=7-4x_3, x_3)$$

Hints :
1/ multiply out the left hand side, this should be a vector. Then equate this with the vector on the right hand side.
Each component has to be the same on LHS and RHS which gives three equations with two unknowns

2/ Compute the inverse of the matrix and left multiply by it. EDIT : I didn't see your equation is not square so forget this hint for your equation.

3/ Use echelon forms

4/ Expect to have a degree of freedom. If this is the case, write $x_2$ in terms of $x_1$ for example.