Discussion
Let \(A = \begin{pmatrix} 0 & 2 \\ 3 & 4 \end{pmatrix}\), \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\). If
\[ (I + A)\begin{pmatrix} 4 & -3 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 8 & -5 \\ 22 & x \end{pmatrix}, \]
then the value of x is equal to
Answer: B
\(x = -14\).
\(I + A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}\). The entry in row 2, column 2 of the product is row 2 of \(I + A\) times column 2 of the other matrix:
\[ x = 3(-3) + 5(-1) = -9 - 5 = -14 \]
Check the given entries: row 1 gives \(1(4) + 2(2) = 8\) and \(1(-3) + 2(-1) = -5\), and row 2, column 1 gives \(3(4) + 5(2) = 22\), all matching. 14 drops the sign, and ±12 forget the identity added to A.
Only one entry of the product is needed, so there is no need to multiply the whole matrices out.
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