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Let A = 1, 2, 3; 0, 4, 5; 1, 0, 6. Find |A|…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Let $A =

\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix}

$. Find$ |A| $. Then find the value of$ a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} $, which multiplies the entries of row$ 1 $by the cofactors of row$ 2$.

Need a reminder? The key ideaShow

For order $2$ , $|A| = ad - bc$ . The minor $M_{ij}$ is the determinant left after deleting row $i$ and column $j$ , and the cofactor is $C_{ij} = (-1)^{i+j}M_{ij}$ . The determinant is the sum of (entry times cofactor) along any row or column.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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