Find x if [x 1] * [[1, 0], [-2, -3]] * [[x], [3]] = 0

Steps to Solve

1. Solving for x in the first equation The given equation is: $[x \ 1] \begin{bmatrix} 1 & 0 \ -2 & -3 \end{bmatrix} \begin{bmatrix} x \ 3 \end{bmatrix} = 0$

First multiply the two matrices in the middle: $\begin{bmatrix} 1 & 0 \ -2 & -3 \end{bmatrix} \begin{bmatrix} x \ 3 \end{bmatrix} = \begin{bmatrix} (1)(x) + (0)(3) \ (-2)(x) + (-3)(3) \end{bmatrix} = \begin{bmatrix} x \ -2x - 9 \end{bmatrix}$

Now, multiply the row matrix with the resulting column matrix : $[x \ 1] \begin{bmatrix} x \ -2x - 9 \end{bmatrix} = (x)(x) + (1)(-2x - 9) = x^2 - 2x - 9$

So, we have the equation $x^2 - 2x - 9 = 0$ Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1, b = -2, c = -9$ $x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-9)}}{2(1)} = \frac{2 \pm \sqrt{4 + 36}}{2} = \frac{2 \pm \sqrt{40}}{2} = \frac{2 \pm 2\sqrt{10}}{2} = 1 \pm \sqrt{10}$ Thus, $x = 1 + \sqrt{10}$ or $x = 1 - \sqrt{10}$

2. Solving for x in the second equation The given equation is: $[2x \ 3] \begin{bmatrix} 1 \ -3 \end{bmatrix} = 0$ Multiply the row matrix by the column matrix: $(2x)(1) + (3)(-3) = 2x - 9$ So, we have the equation $2x - 9 = 0$ $2x = 9$ $x = \frac{9}{2}$

3. Solving for a and b in the third equation The matrix equation is: $\begin{bmatrix} 4 & 2 \ 3 & -1 \end{bmatrix} \begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} -4 \ 2 \end{bmatrix}$ This corresponds to the system of linear equations: $4a + 2b = -4$ $3a - b = 2$ Multiplying the second equation by 2 gives: $6a - 2b = 4$ Adding this to the first equation eliminates $b$: $4a + 2b + 6a - 2b = -4 + 4$ $10a = 0$ $a = 0$ Substituting $a = 0$ into the equation $3a - b = 2$: $3(0) - b = 2$ $-b = 2$ $b = -2$ Thus $a = 0$ and $b = -2$