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Determinants and Matrices — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDeterminants and Matrices3 Marks
Question
If $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 1 \\ 4 & -1 \\ -3 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$ then show that

$(A-C)^{\top}=A^{\top}-C^{\top}$

✓

Answer

$A-C=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]-\left[\begin{array}{cc}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$

$=\left[\begin{array}{cc}2-1 & -3-2 \\ 5+1 & -4-4 \\ -6+2 & 1-3\end{array}\right]$

$=\left[\begin{array}{cc}1 & -5 \\ 6 & -8 \\ -4 & -2\end{array}\right]$

$\therefore \quad(A-C)^T=\left[\begin{array}{ccc}1 & 6 & -4 \\ -5 & -8 & -2\end{array}\right]$

$\ldots(\mathrm{i})$

Now, $A^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 5 & -6 \\ -3 & -4 & 1\end{array}\right]$ and

$C^T=\left[\begin{array}{ccc}1 & -1 & -2 \\ 2 & 4 & 3\end{array}\right]$

$\therefore \quad A^T-C^T=\left[\begin{array}{ccc}2 & 5 & -6 \\ -3 & -4 & 1\end{array}\right]-\left[\begin{array}{ccc}1 & -1 & -2 \\ 2 & 4 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}2-1 & 5+1 & -6+2 \\ -3-2 & -4-4 & 1-3\end{array}\right]$

$=\left[\begin{array}{ccc}1 & 6 & -4 \\ -5 & -8 & -2\end{array}\right]$

...(ii)

From (i) and (ii), we get

$(A-C)^{\top}=A^{\top}-C^{\top}$

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