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Matrices and Determinants question

2023 · 6 Apr · Shift 1 · Q28

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Matrices and Determinants
2023 · 6 Apr · Shift 1 · Q28

Matrices and Determinants question

2023 · 6 Apr · Shift 1 · Q28

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1

Let

\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{2 \times 2}

, where

\mathrm{a}_{\mathrm{ij}} eq 0

for all

\mathrm{i}, \mathrm{j}

and

\mathrm{A}^{2}=\mathrm{I}

. Let a be the sum of all diagonal elements of

\mathrm{A}

and

\mathrm{b}=|\mathrm{A}|

. Then

3 a^{2}+4 b^{2}

is equal to :

A
4
B
3
C
14
D
7

View written solutionFree

Correct answer: A

Let $A=\begin{pmatrix}p&q\\ r&s\end{pmatrix},$ with all entries nonzero, and given $A^2=I.$

We need:

$a$ = sum of diagonal elements = trace of $A$ = $p+s$
$b=|A|=\det(A)$

Then find $3a^2+4b^2$ .

Use the condition $A^2=I$ .

For a $2\times 2$ matrix,

=\begin{pmatrix}p^2+qr & q(p+s)\\ r(p+s) & rq+s^2\end{pmatrix}.

Since $A^2=I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ , we get $p^2+qr=1, \qquad q(p+s)=0, \qquad r(p+s)=0, \qquad rq+s^2=1.$

Given $q\neq 0$ and $r\neq 0$ , from $q(p+s)=0, \quad r(p+s)=0,$ we must have $p+s=0.$

Therefore, $a=p+s=0.$

Now find $b=\det(A)$ .

Since $A^2=I$ , taking determinants on both sides gives $\det(A^2)=\det(I)=1.$ So, $(\det A)^2=1 \implies b^2=1.$

Thus, $b^2=1.$

Compute the required expression: $3a^2+4b^2=3(0)^2+4(1)=4.$

Check options:

A: $4$ ✅
B: $3$
C: $14$
D: $7$

Hence the correct answer is $\boxed{4}.$

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