Let S=A=[array|c|c| a & b c & d array]:a,b,c,d 0,1,2,3,4 and A^2-4A+3I=0 be a set of 2 2 m…

Question

Let $S={A=[\begin{array}{|c|c|} a & b \ c & d \end{array}]:a,b,c,d\in {0,1,2,3,4}$ and ${A}^{2}-4A+3I=0}$ be a set of $2	imes 2$ matrices. Then the number of matrices in S , for which the sum of the diagonal elements is equal to 4 , is :

Accepted Answer

Let

$A=[\begin{gathered} a b \ c d \end{gathered}],\, a,b,c,d\in {0,1,2,3,4}$

and it satisfies

${A}^{2}-4A+3I=0.$

We have to count those matrices for which the sum of diagonal elements is

$a+d=4.$

So we must find the number of matrices satisfying both:

${A}^{2}-4A+3I=0$

$tr(A)=a+d=4$

Factor the matrix equation: We write

${A}^{2}-4A+3I=(A-I)(A-3I)=0.$

So the matrix $A$ satisfies the polynomial

$(x-1)(x-3)=0.$

Hence the only possible eigenvalues are $1$ and $3$.

Now the trace is given to be $4$. Since trace is sum of eigenvalues, the eigenvalues must be

$13.$

So every such matrix has distinct eigenvalues $1$ and $3$, hence it is diagonalizable.

Use trace and determinant: For a $2	imes 2$ matrix with eigenvalues $1$ and $3$,

trace $=1+3=4$

determinant $=1\cdot 3=3$

So we need all matrices with entries in ${0,1,2,3,4}$ such that

$a+d=4,\, ad-bc=3.$

Let $d=4-a$. Then

$a(4-a)-bc=3$

that is,

$bc=a(4-a)-3.$

Now check all possible values of $a\in {0,1,2,3,4}$.

Case-wise counting: Case 1: $a=0$
Then $d=4$, and

$bc=0\cdot 4-3=-3$

not possible.

Case 2: $a=1$
Then $d=3$, and

$bc=1\cdot 3-3=0$

So we need ordered pairs $(b,c)$ with $b,c\in {0,1,2,3,4}$ and $bc=0$.

That means at least one of $b,c$ is $0$.

Number of such ordered pairs:

$b=0$: $5$ choices for $c$

$c=0$: $5$ choices for $b$

but $(0,0)$ counted twice

So total

$5+5-1=9$

matrices.

Case 3: $a=2$
Then $d=2$, and

$bc=2\cdot 2-3=1$

So

$bc=1.$

Since $b,c\in {0,1,2,3,4}$, this is possible only for

$(b,c)=(1,1).$

So total matrices in this case:

$1$

Case 4: $a=3$
Then $d=1$, and

$bc=3\cdot 1-3=0$

Again, number of ordered pairs $(b,c)$ with $bc=0$ is

$9.$

Case 5: $a=4$
Then $d=0$, and

$bc=4\cdot 0-3=-3$

not possible.

Total count: Total number of matrices is

$9+1+9=19.$

Hence the required number of matrices is

$19$

So, the correct option is:

Solutions · 1