AHL 1.14—Introduction to matrices

The matrix $P$ has 2 rows and 3 columns, so its order is $2\times 3$ (rows $\times$ columns).

The notation $p_{ij}$ refers to the element in row $i$, column $j$. So $p_{23}$ is in row 2, column 3.

Row 2 of $P$ is $(0,5,-4)$. The element in column 3 of this row is $-4$.

We need the order (rows $\times$ columns) and the value of $p_{23}$ (row 2, column 3).

Options stating 'Order $3\times 2$' reverse the rows and columns. $P$ has 2 rows and 3 columns, so $3\times 2$ is incorrect. This eliminates 'Order $3\times 2$; $p_{23}=2$' and 'Order $3\times 2$; $p_{23}=-4$'.

Among the remaining options with order $2\times 3$, the option stating $p_{23}=5$ is wrong because 5 is in row 2, column 2, not column 3.

The correct answer is Order $2\times 3$; $p_{23}=-4$, since row 2, column 3 gives $-4$.

$\det (M)=(3)(5)-(-1)(2)=15+2=17$

$\det (N)=(4)(2)-(k)(-3)=8+3k$

$8+3k=17⟹3k=9⟹k=3$

Wait — re-checking: $\det (N)=(4)(2)-(k)(-3)=8+3k$. Setting $8+3k=17$ gives $k=3$. The correct answer is $k=3$.

We need $\det (M)=\det (N)$. First, $\det (M)=(3)(5)-(-1)(2)=17$ and $\det (N)=(4)(2)-(k)(-3)=8+3k$.

$\det (N)=8+3(-3/2)=8-4.5=3.5\ne 17$. Eliminate.

$\det (N)=8+3(3/2)=8+4.5=12.5\ne 17$. Eliminate.

$\det (N)=8+3(-3)=8-9=-1\ne 17$. Eliminate.

Testing $k=3$: $\det (N)=8+9=17=\det (M)$. ✓ The answer is $k=3$.