Question

Let \( B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & -1 & -1 \end{bmatrix} \), \( C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix} \), and if a matrix \( A \) is such that \( BAC = I \), then \( A^{-1} = \)

• A. \( \begin{bmatrix} -3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6 \end{bmatrix} \)
• B. \( \begin{bmatrix} -3 & -5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{bmatrix} \)
• C. \( \begin{bmatrix} -3 & -5 & -6 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix} \)
• D. \( \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix} \)

Hint:

When \( BAC = I \), use \( A^{-1} = CB \) for the solution.

Answer 1

The Correct Option is D

We are given that \( BAC = I \), so: \[ A^{-1} = CB \] Now, compute the matrix product \( CB \): \[ CB = \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix} \] Thus, the correct answer is option (4).

Answer 2

Step 1: Understand the given equation.
We have matrices \( B, A, C \) such that:
\[ B A C = I, \] where \( I \) is the identity matrix.

Step 2: Express \( A \) in terms of \( B \) and \( C \).
Multiply both sides from the left by \( B^{-1} \) and from the right by \( C^{-1} \) (assuming \( B \) and \( C \) are invertible):
\[ A = B^{-1} I C^{-1} = B^{-1} C^{-1}. \]

Step 3: Find \( A^{-1} \).
Since \( A = B^{-1} C^{-1} \), then:
\[ A^{-1} = (B^{-1} C^{-1})^{-1} = C B. \]

Step 4: Compute \( C B \).
Given:
\[ B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & -1 & -1 \end{bmatrix}, \quad C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}. \]
Multiply \( C \) and \( B \):
\[ A^{-1} = C B = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & -1 & -1 \end{bmatrix}. \]

Calculate each element:
- First row:
\[ (-1)(2) + (0)(1) + (1)(-1) = -2 + 0 -1 = -3, \] \[ (-1)(6) + (0)(0) + (1)(-1) = -6 + 0 -1 = -7, \] \[ (-1)(4) + (0)(1) + (1)(-1) = -4 + 0 -1 = -5. \]
- Second row:
\[ (1)(2) + (1)(1) + (3)(-1) = 2 + 1 -3 = 0, \] \[ (1)(6) + (1)(0) + (3)(-1) = 6 + 0 -3 = 3, \] \[ (1)(4) + (1)(1) + (3)(-1) = 4 + 1 -3 = 2. \]
- Third row:
\[ (2)(2) + (0)(1) + (2)(-1) = 4 + 0 -2 = 2, \] \[ (2)(6) + (0)(0) + (2)(-1) = 12 + 0 -2 = 10, \] \[ (2)(4) + (0)(1) + (2)(-1) = 8 + 0 -2 = 6. \]

So,
\[ A^{-1} = \begin{bmatrix} -3 & -7 & -5 \\ 0 & 3 & 2 \\ 2 & 10 & 6 \end{bmatrix}. \]

Step 5: Check the given correct answer.
Given answer is:
\[ \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix}. \] The difference suggests recalculating carefully the multiplication for the second and third columns.

Step 6: Recalculate the second and third columns:
- Second column:
\[ (-1)(6) + (0)(0) + (1)(-1) = -6 + 0 -1 = -7 \quad (\text{already computed}) \]
But given answer has \(-5\), so check carefully:
Actually, original matrix \( B \) second column is \( [6, 0, -1]^T \) and third column is \( [4,1,-1]^T \). Re-check third column multiplication:
For element \( (1,2) \):
\[ (-1)(6) + (0)(0) + (1)(-1) = -6 + 0 -1 = -7, \] Given answer says \(-5\), so possibly the problem states \( BAC = I \) but implies a different order or a typo.

Conclusion:
Following the correct matrix multiplication, the answer closest to the calculation is:
\[ A^{-1} = C B = \begin{bmatrix} -3 & -7 & -5 \\ 0 & 3 & 2 \\ 2 & 10 & 6 \end{bmatrix}. \]

If the given correct answer is:
\[ \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix}, \]
then either the matrices or problem conditions differ, or it is a provided solution.

Hence, theoretically:
\[ \boxed{A^{-1} = C B}. \]