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Question 15:

You are given matrices:

• $R = [3 \quad 1]$
• $Q = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$

We need to find the correct statement about these matrices.

Let's calculate the expressions given in the options:

Option A: $R + Q$

$R + Q = \begin{bmatrix} 3 \quad 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ Matrix addition requires matrices to have the same dimensions, but $R$ is $1 \times 2$ and $Q$ is $2 \times 1$, so this addition is not possible.

Option B: $R - Q^T$

Transpose $Q$, so: $Q^T = \begin{bmatrix} 1 & 3 \end{bmatrix}$ Now subtract: $R - Q^T = \begin{bmatrix} 3 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -2 \end{bmatrix}$ So, Option B is correct.

Let's verify the other options.

Option C: $R + Q^T$

$R + Q^T = \begin{bmatrix} 3 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 4 \end{bmatrix}$ So, Option C is also correct.

Option D: $R - Q^2$

Calculating $Q^2$ doesn't make sense as $Q$ is not a square matrix, so this is incorrect.

Option E: $Q - R^T$

Since $Q$ is a $2 \times 1$ matrix and $R^T$ is a $2 \times 1$ matrix, subtraction should be: $Q - R^T = \begin{bmatrix} 1 \\ 3 \end{bmatrix} - \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \end{bmatrix}$ So, this is incorrect.

The correct answers are B and C.

Question 16:

Matrix $D = \begin{bmatrix} -2 & 6 \\ 1 & 4 \end{bmatrix}$

We need to compute $D^2$: $D^2 = D \times D = \begin{bmatrix} -2 & 6 \\ 1 & 4 \end{bmatrix} \times \begin{bmatrix} -2 & 6 \\ 1 & 4 \end{bmatrix}$

Multiplying these two matrices: $D^2 = \begin{bmatrix} (-2 \times -2) + (6 \times 1) & (-2 \times 6) + (6 \times 4) \\ (1 \times -2) + (4 \times 1) & (1 \times 6) + (4 \times 4) \end{bmatrix}$