Q.If $A = \begin{bmatrix} i & 2i \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2i & i \\ 2 & -3 \end{bmatrix}$, where $\sqrt{-1}=i$, find $A+B$ and $A-B$. Show that $A+B$ is singular. Is $A-B$ singular? Justify your answer.

Given $A = \begin{bmatrix} i & 2i \\ -3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 2i & i \\ 2 & -3 \end{bmatrix}$, with $\sqrt{-1}=i$.

$A+B = \begin{bmatrix} i+2i & 2i+i \\ -3+2 & 2+(-3) \end{bmatrix} = \begin{bmatrix} 3i & 3i \\ -1 & -1 \end{bmatrix}$

$A-B = \begin{bmatrix} i-2i & 2i-i \\ -3-2 & 2-(-3) \end{bmatrix} = \begin{bmatrix} -i & i \\ -5 & 5 \end{bmatrix}$

Determinant of $A+B$: $(3i)(-1)-(3i)(-1) = -3i+3i = 0$, so $A+B$ is singular — as expected, since Row 1 $=3i\times[1,1]$ and Row 2 $=-1\times[1,1]$, i.e. the two rows are proportional.

Determinant of $A-B$: $(-i)(5)-(i)(-5) = -5i+5i = 0$. So $A-B$ is ALSO singular. …

Add and subtract A and B entry-wise, then compute each 2×2 determinant to test singularity; for A−B …

• Assuming that because the question specifically asks 'is A−B singular, justify' the expected answer must be 'non-singular' — but the direct computation shows $|A-B|=0$ too, so honesty about the actual determinant matters more than guessing the 'expected' cont …