Question

If matrices \( A \) and \( B \) are such that \[ A = \begin{bmatrix} 0 & -2 & 3 \\ -2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \] and \( B(I - A) = (I + A) \), then find \( B \).

• A. \[ B = \begin{bmatrix} -1 & \tfrac{2}{3} & \tfrac{2}{3} \\ -2 & \tfrac{5}{3} & -\tfrac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \]
• B. \[ B = \begin{bmatrix} -1 & \tfrac{1}{3} & \tfrac{1}{3} \\ -2 & \tfrac{5}{3} & -\tfrac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \]
• C. \[ B = \begin{bmatrix} -1 & 0 & \tfrac{2}{3} \\ 0 & \tfrac{5}{3} & -\tfrac{10}{3} \\ 2 & 2 & -3 \end{bmatrix} \]
• D. \[ B = \begin{bmatrix} -1 & \tfrac{2}{3} & \tfrac{2}{3} \\ -\tfrac{2}{3} & 1 & \tfrac{5}{3} \\ -2 & 2 & -3 \end{bmatrix} \]

Hint:

To solve matrix equations involving inverses, first isolate the required matrix and then compute inverses carefully.

Answer 1

The Correct Option is A

To find matrix \( B \) such that \( B(I - A) = (I + A) \), we start by identifying the necessary matrices:

Given matrix \( A \),

\( A = \begin{bmatrix} 0 & -2 & 3 \\ -2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \)

The identity matrix \( I \) for a 3x3 matrix is,

\( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

First, compute \( I - A \):

\( I - A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -2 & 3 \\ -2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \)

Calculating this subtraction, we get:

\( I - A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \)

Now, compute \( I + A \):

\( I + A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -2 & 3 \\ -2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \)

Computing this addition, we obtain:

\( I + A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 1 & 1 \\ -1 & 1 & 1 \end{bmatrix} \)

The main equation given is:

\( B(I - A) = I + A \)

Let \( B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix} \).

Multiplying \( B \) with \( (I - A) \), we have:

\( B(I - A) = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix} \begin{bmatrix} 1 & 2 & -3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \)

By carrying out this matrix multiplication, we set it equal to \( I + A \):

\( \begin{bmatrix} (b_{11} + 2b_{12} + b_{13}) & (2b_{11} + b_{12} - b_{13}) & (-3b_{11} - b_{12} + b_{13}) \\ (b_{21} + 2b_{22} + b_{23}) & (2b_{21} + b_{22} - b_{23}) & (-3b_{21} - b_{22} + b_{23}) \\ (b_{31} + 2b_{32} + b_{33}) & (2b_{31} + b_{32} - b_{33}) & (-3b_{31} - b_{32} + b_{33}) \end{bmatrix} = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 1 & 1 \\ -1 & 1 & 1 \end{bmatrix} \)

Solving these equations using simple algebra, we substitute values from the correct option:

Substituting:

\( B = \begin{bmatrix} -1 & \frac{2}{3} & \frac{2}{3} \\ -2 & \frac{5}{3} & -\frac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \)

Verifying multiple elements confirms that this satisfies the equation \( B(I - A) = (I + A) \). Hence, the correct matrix \( B \) is:

\( B = \begin{bmatrix} -1 & \frac{2}{3} & \frac{2}{3} \\ -2 & \frac{5}{3} & -\frac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \)