Question

Let \(A+B=\begin{bmatrix} 4&1&4\\1&4&4 \end{bmatrix}\)and \(B=\begin{bmatrix} 1&0&-2\\-1&3&0 \end{bmatrix}\), then A=

• A. \(\begin{bmatrix} 3&1&2\\0&3&4 \end{bmatrix}\)
• B. \(\begin{bmatrix} 5&1&2\\0&7&4 \end{bmatrix}\)
• C. \(\begin{bmatrix} 3&-1&-2\\2&1&4 \end{bmatrix}\)
• D. \(\begin{bmatrix} 5&1&6\\2&1&4 \end{bmatrix}\)
• E. \(\begin{bmatrix} 3&1&6\\2&1&4 \end{bmatrix}\)

Answer 1

Answer: (E)

We are given the equation \(A + B = \begin{bmatrix} 4 & 1 & 4 \\ 1 & 4 & 4 \end{bmatrix}\) and the matrix \(B = \begin{bmatrix} 1 & 0 & -2 \\ -1 & 3 & 0 \end{bmatrix}\). We need to find matrix \( A \). 

To solve for \( A \), we subtract matrix \( B \) from both sides of the equation:

\(A = \begin{bmatrix} 4 & 1 & 4 \\ 1 & 4 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 0 & -2 \\ -1 & 3 & 0 \end{bmatrix}\)

Subtracting corresponding elements of the matrices:

\(A = \begin{bmatrix} 4-1 & 1-0 & 4 - (-2) \\ 1 - (-1) & 4 - 3 & 4 - 0 \end{bmatrix}\)

\(A = \begin{bmatrix} 3 & 1 & 6 \\ 2 & 1 & 4 \end{bmatrix}\)

The matrix \( A \) is \( \begin{bmatrix} 3 & 1 & 6 \\ 2 & 1 & 4 \end{bmatrix} \).