The task requires finding all values of $a$ that make the determinant of the given matrix $A$ equal to zero. The matrix $A$ is:

$A = \begin{bmatrix} a & 2 \\ 9-7a & a-46 \end{bmatrix}.$

Determinant Formula for a $2 \times 2$ Matrix:

For a general $2 \times 2$ matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix},$ the determinant is given by: $\text{det}(A) = p \cdot s - q \cdot r.$

Applying this Formula:

Here: $\text{det}(A) = a \cdot (a - 46) - 2 \cdot (9 - 7a).$

Expanding the Determinant:

$\text{det}(A) = a(a - 46) - 2(9 - 7a).$ $\text{det}(A) = a^2 - 46a - 18 + 14a.$ $\text{det}(A) = a^2 - 32a - 18.$

Solve $\text{det}(A) = 0$:

$a^2 - 32a - 18 = 0.$

We will solve this quadratic equation using the quadratic formula: $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Here:

• $a = 1$,
• $b = -32$,
• $c = -18$.

Substitute into the quadratic formula: $a = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(1)(-18)}}{2(1)}.$ $a = \frac{32 \pm \sqrt{1024 + 72}}{2}.$ $a = \frac{32 \pm \sqrt{1096}}{2}.$ $a = \frac{32 \pm 2\sqrt{274}}{2}.$ $a = 16 \pm \sqrt{274}.$

Final Answer:

The values of $a$ that make $\text{det}(A) = 0$ are: $a = 16 + \sqrt{274}, \quad a = 16 - \sqrt{274}.$

Would you like a detailed explanation for any part of this solution or assistance in understanding quadratic equations?

5 Related Questions:

1. How do you compute determinants for higher-order matrices?
2. What does it mean for a matrix to have a determinant of 0?
3. Can the quadratic formula be used in all quadratic equations?
4. How do matrix determinants relate to linear independence?
5. What applications involve solving matrices with zero determinants?

Tip:

When solving quadratic equations, double-check the discriminant ($b^2 - 4ac$) to ensure your solution includes all possible real or complex roots.