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Determinants and Matrices — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDeterminants and Matrices4 Marks
Question
If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 3 & 1 & 2\end{array}\right], B=\left[\begin{array}{lll}2 & 1 & -4 \\ 3 & 5 & -2\end{array}\right]$ and $C=\left[\begin{array}{ccc}0 & 2 & 3 \\ -1 & -1 & 0\end{array}\right]$, verify that $(A+2 B+$

$3 C)^{\top}=A^{\top}+2 B^{\top}+3 C^{\top}$.

✓

Answer

$\begin{aligned} & =\left[\begin{array}{lll}1 & 0 & 1 \\ 3 & 1 & 2\end{array}\right]+2\left[\begin{array}{ccc}2 & 1 & -4 \\ 3 & 5 & -2\end{array}\right]+3\left[\begin{array}{ccc}0 & 2 & 3 \\ -1 & -1 & 0\end{array}\right] \\ & =\left[\begin{array}{lll}1 & 0 & 1 \\ 3 & 1 & 2\end{array}\right]+\left[\begin{array}{ccc}4 & 2 & -8 \\ 6 & 10 & -4\end{array}\right]+\left[\begin{array}{ccc}0 & 6 & 9 \\ -3 & -3 & 0\end{array}\right] \\ & =\left[\begin{array}{lll}1+4+0 & 0+2+6 & 1-8+9 \\ 3+6-3 & 1+10-3 & 2-4+0\end{array}\right]\end{aligned}$

$\therefore \quad A+2 B+3 C=\left[\begin{array}{ccc}5 & 8 & 2 \\ 6 & 8 & -2\end{array}\right]$

$\therefore \quad[\mathrm{A}+2 \mathrm{~B}+3 \mathrm{C}]^{\mathrm{T}}=\left[\begin{array}{cc}3 & 0 \\ 8 & 8 \\ 2 & -2\end{array}\right]$

$\ldots(\mathrm{i})$

Now, $A^T=\left[\begin{array}{ll}1 & 3 \\ 0 & 1 \\ 1 & 2\end{array}\right], B^T=\left[\begin{array}{cc}2 & 3 \\ 1 & 5 \\ -4 & -2\end{array}\right]$

and $C^T=\left[\begin{array}{cc}0 & -1 \\ 2 & -1 \\ 3 & 0\end{array}\right]$

$\therefore \quad A^T+2 B^T+3 C^T$

$\begin{aligned} & =\left[\begin{array}{ll}1 & 3 \\ 0 & 1 \\ 1 & 2\end{array}\right]+2\left[\begin{array}{cc}2 & 3 \\ 1 & 5 \\ -4 & -2\end{array}\right]+3\left[\begin{array}{cc}0 & -1 \\ 2 & -1 \\ 3 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 3 \\ 0 & 1 \\ 1 & 2\end{array}\right]+\left[\begin{array}{cc}4 & 6 \\ 2 & 10 \\ -8 & -4\end{array}\right]+\left[\begin{array}{cc}0 & -3 \\ 6 & -3 \\ 9 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}1+4+0 & 3+6-3 \\ 0+2+6 & 1+10-3 \\ 1-8+9 & 2-4+0\end{array}\right]\end{aligned}$

$\therefore A^{\top}+2 B^{\top}+3 C^{\top}=\left[\begin{array}{cc}5 & 6 \\ 8 & 8 \\ 2 & -2\end{array}\right]$

From (i) and (ii), we get

$(A+2 B+3 C)^{\top}=A^{\top}+2 B^{\top}+3 C^{\top}$

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