If $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right],…
If $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 3 \\ 3 & 0 & 4\end{array}\right]$, $C=\left[\begin{array}{lll}2 & 0 & 1 \\ 0 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$, then $\left(\left(\left((A B C)^{-1}\right)^T\right)^{-1}\right)^T=$
- $\left[\begin{array}{ccc}64 & 39 & 28 \\ 29 & 16 & 11 \\ 11 & 2 & 5\end{array}\right]$
- $\left[\begin{array}{ccc}63 & 39 & 20 \\ 29 & 16 & 11 \\ 10 & 2 & 5\end{array}\right]$
- $\left[\begin{array}{ccc}64 & 39 & 27 \\ 28 & 15 & 11 \\ 11 & 2 & 5\end{array}\right]$
- $\left[\begin{array}{ccc}61 & 39 & 28 \\ 29 & 16 & 11 \\ 11 & 0 & 5\end{array}\right]$
Solution
Given matrices $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 3 \\ 3 & 0 & 4\end{array}\right]$
$
\begin{aligned}
& \text { and } C=\left[\begin{array}{lll}
2 & 0 & 1 \\
0 & 1 & 0 \\
3 & 2 & 1
\end{array}\right] \\
& \because\left(\left(\left((A B C)^{-1}\right)^T\right)^{-1}\right)^T=\left(\left(\left(C^{-1} B^{-1} A^{-1}\right)^T\right)^{-1}\right)^T \\
& \quad=\left(\left(\left(A^T\right)^{-1}\left(B^T\right)^{-1}\left(C^T\right)^{-1}\right)^{-1}\right)^T=\left(C^T B^T A^T\right)^T=A B C
\end{aligned}
$
$
\begin{aligned}
& =\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & 1 & 1 \\
1 & -1 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 1 & 0 \\
0 & 1 & 3 \\
3 & 0 & 4
\end{array}\right]\left[\begin{array}{lll}
2 & 0 & 1 \\
0 & 1 & 0 \\
3 & 2 & 1
\end{array}\right] \\
& =\left[\begin{array}{rrr}
10 & 3 & 18 \\
4 & 2 & 7 \\
4 & 0 & 1
\end{array}\right]\left[\begin{array}{lll}
2 & 0 & 1 \\
0 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]=\left[\begin{array}{ccc}
64 & 39 & 28 \\
29 & 16 & 11 \\
11 & 2 & 5
\end{array}\right]
\end{aligned}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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