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: Compute \( 7I \)

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

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

\), so \( 7I = 7

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

=

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

\)

Step2: Compute \( A - 7I \)

Given \( A=

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

\), then \( A - 7I=

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

=

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

\)

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

The transpose of a matrix \(

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\) is \(

$$\begin{bmatrix}a&c\\b&d\end{bmatrix}$$

\), so \( (A - 7I)^T=

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

\)

Step4: Compute \( BC \)

Given \( B=

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

\) and \( C=

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

\)

$$
LATEXBLOCK11
$$

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

$$
LATEXBLOCK12
$$

Answer:

\(

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

\)