If `A=[(0,3),(-7,5)]` and `I=[(1,0),(0,1)]`, then find 'k' so that `k^(2)=5A-21I`.

Answer

To solve the problem step by step, we will follow the instructions given in the question.

### Step 1: Identify the given matrices
We have:
- Matrix \( A = \begin{pmatrix} 0 & 3 \\ -7 & 5 \end{pmatrix} \)
- Identity matrix \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)

### Step 2: Calculate \( 5A \)
To find \( 5A \), we multiply each element of matrix \( A \) by 5:
\[ 5A = 5 \cdot \begin{pmatrix} 0 & 3 \\ -7 & 5 \end{pmatrix} = \begin{pmatrix} 5 \cdot 0 & 5 \cdot 3 \\ 5 \cdot -7 & 5 \cdot 5 \end{pmatrix} = \begin{pmatrix} 0 & 15 \\ -35 & 25 \end{pmatrix} \]

### Step 3: Calculate \( 21I \)
Next, we calculate \( 21I \) by multiplying each element of the identity matrix \( I \) by 21:
\[ 21I = 21 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 21 & 0 \\ 0 & 21 \end{pmatrix} \]

### Step 4: Calculate \( 5A - 21I \)
Now, we subtract \( 21I \) from \( 5A \):
\[ 5A - 21I = \begin{pmatrix} 0 & 15 \\ -35 & 25 \end{pmatrix} - \begin{pmatrix} 21 & 0 \\ 0 & 21 \end{pmatrix} = \begin{pmatrix} 0 - 21 & 15 - 0 \\ -35 - 0 & 25 - 21 \end{pmatrix} = \begin{pmatrix} -21 & 15 \\ -35 & 4 \end{pmatrix} \]

### Step 5: Set \( k^2 \) equal to \( 5A - 21I \)
We have:
\[ k^2 = \begin{pmatrix} -21 & 15 \\ -35 & 4 \end{pmatrix} \]

### Step 6: Calculate the determinant of \( k^2 \)
To find \( k \), we need to calculate the determinant of the matrix \( k^2 \):
\[ \text{det}(k^2) = (-21)(4) - (15)(-35) = -84 + 525 = 441 \]

### Step 7: Find \( k \)
Since \( k^2 = 441 \), we take the square root to find \( k \):
\[ k = \sqrt{441} = \pm 21 \]

### Final Answer
Thus, the value of \( k \) is \( \pm 21 \).

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