The number of matrices A=(\beginarraylla & b c & d\endarray) , where a, b, c, d \in\-1,0,1,2,3, \ldots \ldots, 10\ , such that A=A^-1 , is \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2022 · 26 Jul · Shift 2 · Q39 with Solution

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Correct answer: 50

We need to count all $2\times 2$ matrices $A=\begin{pmatrix}a&b\\ c&d\end{pmatrix},\qquad a,b,c,d\in\{-1,0,1,2,\dots,10\}$ such that $A=A^{-1}.$

This means $A^2=I.$ So we must solve the matrix equation $\begin{pmatrix}a&b\\ c&d\end{pmatrix}^2=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$

1. Compute $A^2$

We have $A^2=\begin{pmatrix}a^2+bc & b(a+d)\\ c(a+d) & d^2+bc\end{pmatrix}.$

Since $A^2=I$ , we get the system

$a^2+bc=1$
$b(a+d)=0$
$c(a+d)=0$
$d^2+bc=1$

From (1) and (4), $a^2=d^2.$ So $d=\pm a.$

Now split into cases.

2. Case I: $a+d\neq 0$

Then from $b(a+d)=0,\qquad c(a+d)=0,$ we must have $b=0,\qquad c=0.$

Then (1) gives $a^2=1,$ so $a=\pm 1.$ Also $d^2=1$ and since $a^2=d^2$ and we are in the case $a+d\neq 0$ , we need $d=a$ .

Thus the possible matrices are

\qquad A=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}.$$ Both entries are allowed since $-1$ is in the set. So this case gives **2 matrices**. --- ## 3. Case II: $a+d=0$ Then $$d=-a.$$ The conditions $b(a+d)=0$ and $c(a+d)=0$ are automatically satisfied. Now equation (1) becomes $$a^2+bc=1.$$ So we need to count all triples $(a,b,c)$ in the allowed set such that $$bc=1-a^2.$$ Also $d=-a$ must belong to the allowed set. Since entries must lie in $$S=\{-1,0,1,2,\dots,10\},$$ we need both $a\in S$ and $-a\in S$. This is possible only for $$a\in\{-1,0,1\}.$$ Now check each value. --- ### (i) $a=-1$ Then $$d=1,$$ and $$bc=1-(-1)^2=1-1=0.$$ So we need $$bc=0.$$ Now $b,c\in S$ where $|S|=12$. Number of ordered pairs $(b,c)$ with $bc=0$: - $b=0$: 12 choices for $c$ - $c=0$: 12 choices for $b$ - subtract double-counted $(0,0)$ once Hence count is $$12+12-1=23.$$ So for $a=-1$, number of matrices = **23**. --- ### (ii) $a=0$ Then $$d=0,$$ and $$bc=1-0=1.$$ So we need $$bc=1.$$ In the set $S$, the only ordered pairs with product $1$ are $$(1,1),\quad (-1,-1).$$ Thus this gives **2 matrices**. --- ### (iii) $a=1$ Then $$d=-1,$$ and again $$bc=1-1=0.$$ So number of ordered pairs $(b,c)$ with $bc=0$ is again $$23.$$ Thus this gives **23 matrices**. --- ## 4. Total count Adding all cases: $$2+23+2+23=50.$$ Therefore, the number of matrices is $$\boxed{50}.$$ --- ## 5. Comparison with stored answer Stored correct answer: $50$ Our derived answer is also $50$, so they agree.

More from Matrices and Determinants

1. Compute

A^2

We have

A^2=\begin{pmatrix}a^2+bc & b(a+d)\\ c(a+d) & d^2+bc\end{pmatrix}.

Since

A^2=I

, we get the system

a^2+bc=1

b(a+d)=0

c(a+d)=0

d^2+bc=1

From (1) and (4),

a^2=d^2.

d=\pm a.

Now split into cases.

2. Case I:

a+d\neq 0

Then from

b(a+d)=0,\qquad c(a+d)=0,

we must have

b=0,\qquad c=0.

Then (1) gives

a^2=1,

a=\pm 1.

Also

d^2=1

and since

a^2=d^2

and we are in the case

a+d\neq 0

, we need

d=a

Thus the possible matrices are

\qquad A=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}.$$ Both entries are allowed since $-1$ is in the set. So this case gives **2 matrices**. --- ## 3. Case II: $a+d=0$ Then $$d=-a.$$ The conditions $b(a+d)=0$ and $c(a+d)=0$ are automatically satisfied. Now equation (1) becomes $$a^2+bc=1.$$ So we need to count all triples $(a,b,c)$ in the allowed set such that $$bc=1-a^2.$$ Also $d=-a$ must belong to the allowed set. Since entries must lie in $$S=\{-1,0,1,2,\dots,10\},$$ we need both $a\in S$ and $-a\in S$. This is possible only for $$a\in\{-1,0,1\}.$$ Now check each value. --- ### (i) $a=-1$ Then $$d=1,$$ and $$bc=1-(-1)^2=1-1=0.$$ So we need $$bc=0.$$ Now $b,c\in S$ where $|S|=12$. Number of ordered pairs $(b,c)$ with $bc=0$: - $b=0$: 12 choices for $c$ - $c=0$: 12 choices for $b$ - subtract double-counted $(0,0)$ once Hence count is $$12+12-1=23.$$ So for $a=-1$, number of matrices = **23**. --- ### (ii) $a=0$ Then $$d=0,$$ and $$bc=1-0=1.$$ So we need $$bc=1.$$ In the set $S$, the only ordered pairs with product $1$ are $$(1,1),\quad (-1,-1).$$ Thus this gives **2 matrices**. --- ### (iii) $a=1$ Then $$d=-1,$$ and again $$bc=1-1=0.$$ So number of ordered pairs $(b,c)$ with $bc=0$ is again $$23.$$ Thus this gives **23 matrices**. --- ## 4. Total count Adding all cases: $$2+23+2+23=50.$$ Therefore, the number of matrices is $$\boxed{50}.$$ --- ## 5. Comparison with stored answer Stored correct answer: $50$ Our derived answer is also $50$, so they agree.

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