KCET · Maths · Matrices
If \(A=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right] B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]\), then \((A B)^{\prime}\) is equal to
- A \(\left[\begin{array}{cc}-3 & -2 \\ 10 & 7\end{array}\right]\)
- B \(\left[\begin{array}{cc}-3 & 10 \\ -2 & 7\end{array}\right]\)
- C \(\left[\begin{array}{ll}-3 & 7 \\ 10 & 2\end{array}\right]\)
- D \(\left[\begin{array}{cc}-3 & 7 \\ 10 & -2\end{array}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\begin{array}{cc}-3 & 10 \\ -2 & 7\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Given, \(A=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right], B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]\)
\(\begin{aligned}
A B &=\left[\begin{array}{ccc}
1 & -2 & 1 \\
2 & 1 & 3
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right] \\
&=\left[\begin{array}{cc}
2-6+1 & 1-4+1 \\
4+3+3 & 2+2+3
\end{array}\right] \\
&=\left[\begin{array}{cc}
-3 & -2 \\
10 & 7
\end{array}\right] \\
A B^{\prime} &=\left[\begin{array}{cc}
-3 & 10 \\
-2 & 7
\end{array}\right]
\end{aligned}\)
\(\begin{aligned}
A B &=\left[\begin{array}{ccc}
1 & -2 & 1 \\
2 & 1 & 3
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right] \\
&=\left[\begin{array}{cc}
2-6+1 & 1-4+1 \\
4+3+3 & 2+2+3
\end{array}\right] \\
&=\left[\begin{array}{cc}
-3 & -2 \\
10 & 7
\end{array}\right] \\
A B^{\prime} &=\left[\begin{array}{cc}
-3 & 10 \\
-2 & 7
\end{array}\right]
\end{aligned}\)
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