The number of matrices A=begin{bmatrix} a & b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $A = A^{-1}$.This implies $A^2 = I$,so $\begin{bmatrix} a & b \ c & d \end{bmatrix}

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $A = A^{-1}$.This implies $A^2 = I$,so $\begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.This gives the system of equations:$1) a^2 + bc = 1$$2) b(a + d) = 0$$3) c(a + d) = 0$$4) bc + d^2 = 1$From $(1)$ and $(4)$,$a^2 = d^2$,so $a = d$ or $a = -d$.Case $I$: $a = -d$. Then $b(a - a) = 0$ is always true. We need $a^2 + bc = 1$.If $a = 0$,then $d = 0$ and $bc = 1$. Since $b, c \in \{-1, 0, 1, \ldots, 10\}$,$bc = 1$ implies $(b, c) = (1, 1)$ or $(-1, -1)$. ($2$ pairs).If $a = 1$,then $d = -1$ and $1 + bc = 1 \Rightarrow bc = 0$. This means $b=0$ ($12$ values for $c$) or $c=0$ ($12$ values for $b$). Excluding $(0,0)$ which is counted twice,we have $12 + 12 - 1 = 23$ pairs.If $a = -1$,then $d = 1$ and $1 + bc = 1 \Rightarrow bc = 0$. Similarly,$23$ pairs.Case $II$: $a = d$. Then $b(2a) = 0$ and $c(2a) = 0$. If $a 
eq 0$,then $b = c = 0$. Since $a^2 = 1$,$a = 1$ or $a = -1$. This gives $(1, 1)$ and $(-1, -1)$ for $(a, d)$. ($2$ pairs).If $a = 0$,then $d = 0$,which leads to $bc = 1$,already covered in Case $I$.Total = $2 + 23 + 23 + 2 = 50$.

The number of matrices $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,where $a, b, c, d \in \{-1, 0, 1, 2, 3, \ldots, 10\}$,such that $A=A^{-1}$,is

Similar Questions

If $\omega$ is a root of the equation $x+\frac{1}{x}+1=0$,then the value of the determinant $\left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \\ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \\ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array}\right|$ is equal to

If $\left|\begin{array}{ccc}-1 & 7 & 0 \\ 2 & 1 & -3 \\ 3 & 4 & 1\end{array}\right|=A$, then the value of $\left|\begin{array}{ccc}13 & -11 & 5 \\ -7 & -1 & 25 \\ -21 & -3 & -15\end{array}\right|$ is:

$A$ is a $3 \times 3$ matrix satisfying $A^3-5A^2+7A+I=0$. If $A^5-6A^4+12A^3-6A^2+2A+2I=lA+mI$, then $l+m=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module