If A= [ cc 2 & -1 3 & -2 4 & 1 ] and B= [ ccc 0 & 3 & -4 2 & -1 & 1 ], verify thati. (A B)^ =B^ A^ ii. (BA)^T = A^TB^T

A= [ cc 2 & -1 3 & -2 4 & 1 ] and B= [ ccc 0 & 3 & -4 2 & -1 & 1 ] A^ = [ ccc 2 & 3 & 4 -1 & -2 & 1 ] and B^ T = [ cc 0 & 2 3 & -1 -4 & 1 ] i. AB= [ cc 2 & -1 3 & -2 4 & 1 ] [ ccc 0 & 3 & -4 2 & -1 & 1 ] = [ ccc 0-2 & 6+1 & -8-1 0-4 & 9+2 & -12-2 0+2 & 12-1 & -16+1 ] = [ ccc -2 & 7 & -9 -4 & 11 & -14 2 & 11 & -15 ] (AB)^ T = [ ccc -2 & -4 & 2 7 & 11 & 11 -9 & -14 & -15 ] ..i) B^ T A^ T = [ cc 0 & 2 3 & -1 -4 & 1 ] [ ccc 2 & 3 & 4 -1 & -2 & 1 ] = [ ccc 0-2 & 0-4 & 0+2 6+1 & 9+2 & 12-1 -8-1 & -12-2 & -16+1 ] = [ ccc -2 & -4 & 2 7 & 11 & 11 -9 & -14 & -15 ] ...(ii) From (i) and (ii), we get (A B)^ =B^ A^ ii. BA & = [ ccc 0 & 3 & -4 2 & -1 & 1 ] [ cc 2 & -1 3 & -2 4 & 1 ] & = [ cc 0+9-16 & 0-6-4 4-3+4 & -2+2+1 ] BA= [ cc -7 & -10 5 & 1 ] (BA)^ T = [ cc -7 & 5 -10 & 1 ] A^ T B^ T & = [ ccc 2 & 3 & 4 -1 & -2 & 1 ] [ cc 0 & 2 3 & -1 -4 & 1 ] & = [ cc 0+9-16 & 4-3+4 0-6-4 & -2+2+1 ] & = [ cc -7 & 5 -10 & 1 ] . From (i) and (ii) we get (B A)^ =A^ B^

If A= [ cc 2 & -1 3 & -2 4 & 1 ] and B= [ ccc 0 & 3 & -4 2 & -1 &…

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Question

If $A=\left[\begin{array}{cc}2 & -1 \\ 3 & -2 \\ 4 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}0 & 3 & -4 \\ 2 & -1 & 1\end{array}\right]$, verify thati. $(A B)^{\top}=B^{\top} A^{\top}$
ii. $(BA)^T = A^TB^T$

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Answer

$A=\left[\begin{array}{cc}2 & -1 \\ 3 & -2 \\ 4 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}0 & 3 & -4 \\ 2 & -1 & 1\end{array}\right]$$\therefore \quad A^{\top}=\left[\begin{array}{ccc}2 & 3 & 4 \\ -1 & -2 & 1\end{array}\right]$ and $B^{\mathrm{T}}=\left[\begin{array}{cc}0 & 2 \\ 3 & -1 \\ -4 & 1\end{array}\right]$
i. $\mathrm{AB}=\left[\begin{array}{cc}2 & -1 \\ 3 & -2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ccc}0 & 3 & -4 \\ 2 & -1 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}0-2 & 6+1 & -8-1 \\ 0-4 & 9+2 & -12-2 \\ 0+2 & 12-1 & -16+1\end{array}\right]$
$=\left[\begin{array}{ccc}-2 & 7 & -9 \\ -4 & 11 & -14 \\ 2 & 11 & -15\end{array}\right]$
$\therefore \quad(\mathrm{AB})^{\mathrm{T}}=\left[\begin{array}{ccc}-2 & -4 & 2 \\ 7 & 11 & 11 \\ -9 & -14 & -15\end{array}\right]$
$\ldots$..i)
$\mathbf{B}^{\mathrm{T}} \mathbf{A}^{\mathrm{T}}=\left[\begin{array}{cc}0 & 2 \\ 3 & -1 \\ -4 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 3 & 4 \\ -1 & -2 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}0-2 & 0-4 & 0+2 \\ 6+1 & 9+2 & 12-1 \\ -8-1 & -12-2 & -16+1\end{array}\right]$
$=\left[\begin{array}{ccc}-2 & -4 & 2 \\ 7 & 11 & 11 \\ -9 & -14 & -15\end{array}\right]$
...(ii)
From (i) and (ii), we get
$(A B)^{\top}=B^{\top} A^{\top}$
ii.
$
\begin{aligned}
\mathrm{BA} & =\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
0+9-16 & 0-6-4 \\
4-3+4 & -2+2+1
\end{array}\right]
\end{aligned}
$
$
\therefore \quad \mathrm{BA}=\left[\begin{array}{cc}
-7 & -10 \\
5 & 1
\end{array}\right]
$
$
\therefore \quad(\mathrm{BA})^{\mathrm{T}}=\left[\begin{array}{cc}
-7 & 5 \\
-10 & 1
\end{array}\right]
$
$
\begin{aligned}
A^{\mathrm{T}} B^{\mathrm{T}} & =\left[\begin{array}{ccc}
2 & 3 & 4 \\
-1 & -2 & 1
\end{array}\right]\left[\begin{array}{cc}
0 & 2 \\
3 & -1 \\
-4 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
0+9-16 & 4-3+4 \\
0-6-4 & -2+2+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
-7 & 5 \\
-10 & 1
\end{array}\right] .
\end{aligned}
$
From (i) and (ii) we get
$
(B A)^{\top}=A^{\top} B^{\top}
$

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