Gujarati
3 and 4 .Determinants and Matrices
hard

Suppose $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is a real matrix with non-zero entries, $\alpha d-b c=0$ and $A^2=A$. Then, $a + d$ equals

A

$1$

B

$2$

C

$3$

D

$4$

(KVPY-2018)

Solution

(a)

We have

$\begin{aligned} A &=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ A^2 &=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ A^2 &=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+c d & b c+d^2\end{array}\right] \end{aligned}$

Given, $A^2=A$ and $a d-b c=0$

$\therefore\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+c d & b c+d^2 \end{array}\right]=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$

$a b+b d =b$

$b(a+d) =b$

$a+d =1$

Standard 12
Mathematics

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