Question

For matrix \(A = \begin{bmatrix}3&1\\ 7&5\end{bmatrix}\), the values of x and y so that \(A^2+xI=yA\) are:

• A. x=6, y=8
• B. x = 8,y=6
• C. x=8, y=8
• D. x=6, y=6

Answer 1

The Correct Option is C

To find the values of \(x\) and \(y\) such that \(A^2 + xI = yA\) for the matrix \(A = \begin{bmatrix}3 & 1 \\ 7 & 5\end{bmatrix}\), we proceed as follows:

1. Calculate \(A^2\):
   We have \(A = \begin{bmatrix}3 & 1 \\ 7 & 5\end{bmatrix}\).
   Calculate \(A^2 = A \cdot A = \begin{bmatrix}3 & 1 \\ 7 & 5\end{bmatrix} \cdot \begin{bmatrix}3 & 1 \\ 7 & 5\end{bmatrix}\):
   \(\begin{bmatrix}(3 \times 3) + (1 \times 7) & (3 \times 1) + (1 \times 5) \\ (7 \times 3) + (5 \times 7) & (7 \times 1) + (5 \times 5)\end{bmatrix} = \begin{bmatrix}16 & 8 \\ 46 & 32\end{bmatrix}\).
2. Formulate the equation:
   The identity matrix \(I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\). Therefore, \(xI = x\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}x & 0 \\ 0 & x\end{bmatrix}\).
   We equate \(A^2 + xI\) to \(yA\):
   \(\begin{bmatrix}16 & 8 \\ 46 & 32\end{bmatrix} + \begin{bmatrix}x & 0 \\ 0 & x\end{bmatrix} = \begin{bmatrix}3y & y \\ 7y & 5y\end{bmatrix}\).
   This gives us two matrices that are equal if corresponding elements are equal.
3. Solve for \(x\) and \(y\):
   Equating corresponding elements, we have:
   • \(16 + x = 3y\)
   • \(8 = y\)
   • \(46 = 7y\)
   • \(32 + x = 5y\)
   From \(8 = y\), we substitute \(y = 8\) into the equations:
   • \(16 + x = 3(8) \Rightarrow 16 + x = 24 \Rightarrow x = 8\)
   • \(32 + x = 5(8) \Rightarrow 32 + x = 40 \Rightarrow x = 8\) confirms consistency.

Thus, the values are \(x = 8\) and \(y = 8\).