Question

1. let $a = \

$$\begin{bmatrix} 6 & -4 \\\\ 1 & 3 \\end{bmatrix}$$

$, $b = \

$$\begin{bmatrix} 2 & -5 \\\\ 7 & 4 \\end{bmatrix}$$

$ and $c = \

$$\begin{bmatrix} 3 & 5 \\\\ -4 & 1 \\end{bmatrix}$$

$. compute $(a - 7i)^t - bc$.

Explanation:

Step1: Find \( 7I \)

The identity matrix \( I \) for \( 2\times2 \) is \(

$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

\), so \( 7I =

$$\begin{bmatrix}7&0\\0&7\end{bmatrix}$$

\)

Step2: Compute \( A - 7I \)

\( A=

$$\begin{bmatrix}6&-4\\1&3\end{bmatrix}$$

\), so \( A - 7I=

$$\begin{bmatrix}6 - 7&-4 - 0\\1 - 0&3 - 7\end{bmatrix}$$

=

$$\begin{bmatrix}-1&-4\\1&-4\end{bmatrix}$$

\)

Step3: Find \( (A - 7I)^T \)

Transpose of \(

$$\begin{bmatrix}-1&-4\\1&-4\end{bmatrix}$$

\) is \(

$$\begin{bmatrix}-1&1\\-4&-4\end{bmatrix}$$

\)

Step4: Compute \( BC \)

\( B=

$$\begin{bmatrix}2&-5\\7&4\end{bmatrix}$$

\), \( C=

$$\begin{bmatrix}3&5\\-4&1\end{bmatrix}$$

\)
\( BC=

$$\begin{bmatrix}2\times3+(-5)\times(-4)&2\times5+(-5)\times1\\7\times3+4\times(-4)&7\times5+4\times1\end{bmatrix}$$

=

$$\begin{bmatrix}6 + 20&10 - 5\\21 - 16&35 + 4\end{bmatrix}$$

=

$$\begin{bmatrix}26&5\\5&39\end{bmatrix}$$

\)

Step5: Compute \( (A - 7I)^T - BC \)

\(

$$\begin{bmatrix}-1&1\\-4&-4\end{bmatrix}$$

-

$$\begin{bmatrix}26&5\\5&39\end{bmatrix}$$

=

$$\begin{bmatrix}-1 - 26&1 - 5\\-4 - 5&-4 - 39\end{bmatrix}$$

=

$$\begin{bmatrix}-27&-4\\-9&-43\end{bmatrix}$$

\)

Answer:

\(

$$\begin{bmatrix}-27&-4\\-9&-43\end{bmatrix}$$

\)