Find x if \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} x \\ 3 \end{bmatrix} = 0
Understand the Problem
The question asks to solve for x in the given matrix equation. We are given a matrix equation involving a 1x2 matrix, a 2x2 and a 2x1 matrix, which equals the zero vector. By performing matrix multiplication, we can produce two equations, which we can then solve for x.
Answer
$x = 1 \pm \sqrt{10}$
Answer for screen readers
$x = 1 \pm \sqrt{10}$
Steps to Solve
- Multiply the first two matrices
Multiply the 1x2 matrix $[x \quad 1]$ by the 2x2 matrix $\begin{bmatrix}1 & 0 \ -2 & -3\end{bmatrix}$.
$[x \quad 1]\begin{bmatrix} 1 & 0 \ -2 & -3 \end{bmatrix} = [x(1) + 1(-2) \quad x(0) + 1(-3)] = [x - 2 \quad -3]$
- Multiply the resulting matrix by the last matrix
Multiply the 1x2 matrix $[x-2 \quad -3]$ by the 2x1 matrix $\begin{bmatrix} x \ 3 \end{bmatrix}$.
$[x-2 \quad -3]\begin{bmatrix} x \ 3 \end{bmatrix} = (x-2)(x) + (-3)(3) = x^2 - 2x - 9$
- Set the result equal to zero and solve for x
Since the entire expression is equal to zero we have:
$x^2 - 2x - 9 = 0$
Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -2$, and $c = -9$.
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-9)}}{2(1)} = \frac{2 \pm \sqrt{4 + 36}}{2} = \frac{2 \pm \sqrt{40}}{2} = \frac{2 \pm 2\sqrt{10}}{2} = 1 \pm \sqrt{10}$
Therefore, $x = 1 + \sqrt{10}$ or $x = 1 - \sqrt{10}$.
$x = 1 \pm \sqrt{10}$
More Information
The solutions are $x = 1 + \sqrt{10} \approx 4.162$ and $x = 1 - \sqrt{10} \approx -2.162$.
Tips
A common mistake is made during matrix multiplication by multiplying the wrong elements. Also, forgetting the quadratic formula or messing up the values of a, b, and c will lead to an incorrect answer.
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