Search any question & find its solution
Question:
Answered & Verified by Expert
If $A^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$ then verify that
(i) $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$
(ii) $(\mathbf{A}-\mathbf{B})^{\prime}=\mathbf{A}^{\prime}-\mathbf{B}^{\prime}$
(i) $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$
(ii) $(\mathbf{A}-\mathbf{B})^{\prime}=\mathbf{A}^{\prime}-\mathbf{B}^{\prime}$
Solution:
2324 Upvotes
Verified Answer
(i) $\mathrm{A}^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 1 \\ 0 & 2\end{array}\right] \Rightarrow \mathrm{A}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]$
$\mathrm{A}+\mathrm{B}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$
$=\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]$
$\therefore(A+B)^{\prime}=\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]^{\prime}=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$
R.H.S $=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$
$=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]=$ L.H.S.
Hence, $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$.
(i) $A^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$
$\therefore \mathrm{A}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]$;
$A-B=\left[\begin{array}{rrr}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]$
L.H.S. $=(\mathrm{A}-\mathrm{B})^{\prime}=\left[\begin{array}{rrr}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
R.H.S. $=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$
$=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
Hence, $(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$.
$\mathrm{A}+\mathrm{B}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$
$=\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]$
$\therefore(A+B)^{\prime}=\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]^{\prime}=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$
R.H.S $=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$
$=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]=$ L.H.S.
Hence, $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$.
(i) $A^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$
$\therefore \mathrm{A}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]$;
$A-B=\left[\begin{array}{rrr}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]$
L.H.S. $=(\mathrm{A}-\mathrm{B})^{\prime}=\left[\begin{array}{rrr}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
R.H.S. $=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$
$=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$
Hence, $(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.