Question

If \(A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}, \; B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}\) and \((A + B)^2 = A^2 + B^2\), then the values of \(a\) and \(b\) are:

• A. \(a = 4, \; b = 1\)
• B. \(a = 1, \; b = 4\)
• C. \(a = 0, \; b = 4\)
• D. \(a = 2, \; b = 4\)

Hint:

Always expand matrix equations carefully. Here, the key trick was recognizing that \((A+B)^2 = A^2 + B^2\) implies \(AB + BA = 0\).

Answer 1

The Correct Option is B

Step 1: Expand the matrix condition. \[ (A + B)^2 = A^2 + B^2 \] \[ A^2 + AB + BA + B^2 = A^2 + B^2 \] \[ \Rightarrow AB + BA = 0 \]

Step 2: Compute \(AB\). \[ AB = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \] \[ = \begin{bmatrix} 1 \cdot a + (-1)\cdot b & 1\cdot 1 + (-1)(-1) \\ 2a + (-1)b & 2 \cdot 1 + (-1)(-1) \end{bmatrix} \] \[ = \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} \]

Step 3: Compute \(BA\). \[ BA = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \] \[ = \begin{bmatrix} a \cdot 1 + 1\cdot 2 & a(-1) + 1(-1) \\ b\cdot 1 + (-1)\cdot 2 & b(-1) + (-1)(-1) \end{bmatrix} \] \[ = \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & -b + 1 \end{bmatrix} \]

Step 4: Condition \(AB + BA = 0\). \[ AB + BA = \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} + \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & -b + 1 \end{bmatrix} \] \[ = \begin{bmatrix} (a - b + a + 2) & (2 - a - 1) \\ (2a - b + b - 2) & (3 - b + 1) \end{bmatrix} \] \[ = \begin{bmatrix} 2a - b + 2 & 1 - a \\ 2a - 2 & 4 - b \end{bmatrix} \] For zero matrix: \[ 2a - b + 2 = 0, 1 - a = 0, 2a - 2 = 0, 4 - b = 0 \]

Step 5: Solve equations. From \(1 - a = 0 \Rightarrow a = 1\). \\ From \(4 - b = 0 \Rightarrow b = 4\). Check: \[ 2a - b + 2 = 2(1) - 4 + 2 = 0 \checkmark \] \[ 2a - 2 = 2(1) - 2 = 0 \checkmark \] Hence solution: \[ a = 1, \; b = 4 \] \[ \boxed{a = 1, \; b = 4} \]