Determine the values of $a$ and $b$ so that the following matrices are singular: (i) $A = \begin{bmatrix} 7 & 3 \\ -2 & a \end{bmatrix}$ (ii) $B = \begin{bmatrix} b-1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{bmatrix}$

Determine the values of a and b so that the following matrices are singular: (i) A = 7 3 \-2 a (ii) B = b-1 2

Exercise 7.4 · Q4

a

(i)

A = \begin{bmatrix} 7 & 3 \\ -2 & a \end{bmatrix}

(ii)

B = \begin{bmatrix} b-1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{bmatrix}

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Impose $|A|=0$ (singularity) on each matrix and solve: $a=-\tfrac67$ for (i) and $b=\tfrac{49}{8}$ for (ii).

A square matrix is singular precisely when its determinant vanishes, so for each matrix we write $|A|=0$ and solve for the unknown.

Step 1 — matrix (i). For $A=\begin{bmatrix}7&3\\-2&a\end{bmatrix}$ , the order-2 determinant is

|A| = 7a - 3(-2) = 7a + 6.

Setting $|A|=0$ : $7a+6=0 \Rightarrow a = -\tfrac{6}{7}$ .

Step 2 — matrix (ii). For $B=\begin{bmatrix}b-1&2&3\\3&1&2\\1&-2&4\end{bmatrix}$ , expand along row 1: …