If \( (AB)^{T}=C \), where \( A=\begin{bmatrix} x & -1\\ 2 & 3\end{bmatrix} \), \( B=\begin{bmatrix} 4 & 5\\ 1 & -2\end{bmatrix} \) and \( C=\begin{bmatrix} 3 & 11\\ 7 & y\end{bmatrix} \), then \( x+y \) is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5
Multiplying row by column:
\( \displaystyle AB=\begin{bmatrix} x & -1\\ 2 & 3\end{bmatrix}\begin{bmatrix} 4 & 5\\ 1 & -2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix} 4x-1 & 5x+2\\ 8+3 & 10-6\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix} 4x-1 & 5x+2\\ 11 & 4\end{bmatrix} \)
The transpose interchanges rows and columns:
\( \displaystyle (AB)^{T}=\begin{bmatrix} 4x-1 & 11\\ 5x+2 & 4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix} 3 & 11\\ 7 & y\end{bmatrix} \)
Equal matrices have equal corresponding entries:
\( 4x-1=3\ \Rightarrow\ x=1 \)
\( y=4 \)
The remaining entry agrees: 5x + 2 = 7 also gives x = 1.
So x + y = 1 + 4 = 5.