MCQ
If $\text{A}=\begin{bmatrix}1&-1\\2&-1\end{bmatrix},\text{B}=\begin{bmatrix}\text{a}&1\\\text{b}&-1\end {bmatrix}$ and $(A + B)^2 = A^2 + B^2$, values of $a$ and $b$ are:
- A$a = 4, b = 1$
- ✓$a = 1, b = 4$
- C$a = 0, b = 4$
- D$a = 2, b = 4$
✓
Answer
Correct option: B.
$a = 1, b = 4$
Here,
$(\text{A+B})^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{A}^2+\text{AB}+\text{BA}+\text{B}^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{AB}+\text{BA}=0$
$\Rightarrow\text{AB}=-\text{BA}$
$\Rightarrow\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}=-\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=-\begin{bmatrix}\text{a}+2&-\text{a}-1\\\text{b}-2&-\text{b}+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=\begin{bmatrix}-\text{a}-2&\text{a}+1\\\text{b}+2&\text{b}-1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\Rightarrow\text{a}+1=2$ and $b-1=3$
$\therefore\ \text{a}=1$ and $b=4$
$(\text{A+B})^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{A}^2+\text{AB}+\text{BA}+\text{B}^2=\text{A}^2+\text{B}^2$
$\Rightarrow\text{AB}+\text{BA}=0$
$\Rightarrow\text{AB}=-\text{BA}$
$\Rightarrow\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}=-\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=-\begin{bmatrix}\text{a}+2&-\text{a}-1\\\text{b}-2&-\text{b}+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=\begin{bmatrix}-\text{a}-2&\text{a}+1\\\text{b}+2&\text{b}-1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\Rightarrow\text{a}+1=2$ and $b-1=3$
$\therefore\ \text{a}=1$ and $b=4$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The feasible solution for a LPP is shown in Figure Let $z=3 x-4 y$ be the objective function. (Maximum value of $z+$ Minimum value of $z$ ) is equal to $....$
→Let $f: R \rightarrow R$ be given by $f(x)=x^2-3$. Then, $f^{-1}$ is given by:
→Out of the following which one is not true
The vectors $a, b $ and $ a + b $ are
→Let $\mathrm{f}: R \rightarrow R$ be function defined as
→$f ( x )=\left\{\begin{array}{cc}3\left(1-\frac{| x |}{2}\right) & \text { if }| x | \leq 2 \text { } \\ 0 & \text { if }| x |>2 \text { }\end{array}\right.$ Let $g: R \rightarrow R$ be given by $g(x)=f(x+2)-f(x-2)$. If $n$ and $m$ denote the number of points in $R$ where $\mathrm{g}$ is not continuous and not differentiable, respectively, then $\mathrm{n}+\mathrm{m}$ is equal to $....$
If $f(x) = \frac{x}{{\sqrt {1 + {x^2}} }}$, then $(fofof)(x) = $
→$\cos \left[ {2{{\cos }^{ - 1}}\frac{1}{5} + {{\sin }^{ - 1}}\frac{1}{5}} \right] = $
→The area common to the parabola $y=2 x^2$ and $y=x^2+4$ is:
→If $A_n$ be the area bounded by the curve $y = (\tan x)^n$ and the lines $x = 0, y = 0$ and $\text{x}=\frac{\pi}{4},$ then for $x > 2$
→If $\text{f(x)}=\begin{cases}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ and f(x) is continous at x = 0, then the value of k is:
→