Question
If $A =\left[\begin{array}{cc}1 & 2 \\ -1 & -2\end{array}\right], B =\left[\begin{array}{cc}2 & a \\ -1 & b\end{array}\right]$ and if $( A + B )^2= A ^2+ B ^2$, find values of $a$ and $b$.

Answer

$
\begin{aligned}
& ( A + B )^2= A ^2+ B ^2 \\
& \therefore( A + B )( A + B )= A ^2+ B ^2 \\
& \therefore A ^2+ AB + BA + B ^2= A ^2+ B ^2 \\
& \therefore AB + BA =0 \\
& \therefore AB =- BA \\
& \therefore\left[\begin{array}{rr}
1 & 2 \\
-1 & -2
\end{array}\right]\left[\begin{array}{rr}
2 & a \\
-1 & b
\end{array}\right]=-\left[\begin{array}{rr}
2 & a \\
-1 & b
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
-1 & -2
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
2-2 & a+2 b \\
-2+2 & -a-2 b
\end{array}\right]=-\left[\begin{array}{rr}
2-a & 4-2 a \\
-1-b & -2-2 b
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
0 & a+2 b \\
0 & -a-2 b
\end{array}\right]=\left[\begin{array}{rr}
a-2 & 2 a-4 \\
1+b & 2+2 b
\end{array}\right]
\end{aligned}
$
By the equality of matrices, we get
$
\begin{aligned}
& 0=a-2 \ldots \ldots . .(1) \\
& 0=1+b \ldots \ldots . .(2) \\
& a+2 b=2 a-4 . . \\
& -a-2 b=2+2 b
\end{aligned}
$
From equations (1) and (2), we get $a=2$ and $b=-1$
The values of $a$ and $b$ satisfy equations (3) and (4) also. Hence, $a=2$ and $b=-1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

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 that
(i) $(A B)^{\top}=B^{\top} A^{\top}$
(ii) $(B A)^{\top}=A^{\top} B^{\top}$
Let $X \sim B(n, p)$. If $n=10$ and $E(X)=5$, using the following activity find $p$ and $\operatorname{Var}(X)$
Solution: $E ( X )=\square=5 \square p =\square, q =\square$
$\operatorname{Var}(X)=\square$
A company manufactures 2 types of goods P and Q that requires copper and brass. Each unit of type P requires 2 grams of brass and 1 gram of copper while one unit of type Q requires 1 gram of brass and 2 grams of copper. The company has only 90 grams of brass and 80 grams of copper. Each unit of types P and Q brings profit of ₹ 400 and ₹ 500 respectively. Find the number of units of each type the company should produce to maximize its profit
Solve the following problem of sequencing for minimizing the total elapsed time and idle time for both the machines.
Job P Q R S T U
$M_1$ 1 4 6 3 5 2
$M_2$ 3 6 8 8 1 5
The optimal sequence of the jobs as follows:
           
Total elasped time is obtained as follows:
Job
sequence
Machine A Machine B Idle time for
Machine B
Time
In
Time
Out
Time
In
Time
Out
P $\square$ 1 1 $\square$ $\square$
U $\square$ 3 4 $\square$ $\square$
S $\square$ 6 9 $\square$ $\square$
Q $\square$ 10 17 $\square$ $\square$
R $\square$ 16 23 $\square$ $\square$
T $\square$ 21 31 $\square$ $\square$
Total elapsed time $T=\square$ minutes
Idle time for Machine $A=T-\square=\square$ minutes
Idle time for Machine $B=\square$
The p.d.f. of a continuous r.v.
$X$ is $ f(x)=\left\{\begin{array}{ll} \frac{3 x^2}{8} & 0Determine the c.d.f. of $X$ and hence find $P(X>0)$
A bill of $\text{₹} 8,000$ drawn on $5^{\text {th }}$ January 2019 for 8 months was discounted for $\text{₹} 7,680$ on a certain date. Find the date on which it was discounted at $10 \%$ p.a.
Solution: Banker's Discount
$\text { B.D. }=\text { F.V. }- \text { C.V. }=8,000-7,680=\text{₹} 320$
Date of drawing $=5^{\text {th }}$ January 2019
$\text { Period }=8 \text { months }$
Nominal due date $=\square$
Legal due date $=\square$
B.D. $=$ Interest on F.V. for $n$ at $10 \%$ p.a.
$ \therefore \text { B.D. }=\frac{\text { F.V. } \times \frac{ n }{365} \times r }{100}$
$\therefore 320=\frac{8000 \times \frac{ n }{365} \times 10}{100}$
$\therefore n =\square \text { days } $
April May June July Aug Sep Total
$15$ $31$ $30$ $31$ $31$ $\square$ $\square$
 
$\therefore$ Date of discounting is
Solve the following differential equation
$y \log y \frac{ d x}{ d y}+x=\log y$
Solve the following equations by method of inversion : $4x – 3y – 2 = 0, 3x – 4y + 6 = 0$
For annuity due,
$C=\text{₹} 20,000, n=3, I=0.1,(1.1)^{-3}=0.7513$
Therefore, $P=\frac{\square}{0.1} \times\left[1-(1+0.1)^{\square}\right]$
$ =2,00,000[1-0.7513]$
$=\text{₹} \square $
Evaluate $\int_0^{ a } x^2( a -x)^{\frac{3}{2}} d x$