Exercise 3.18 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.18$
Question 1.
If $\mathrm{A}$ is of order $\mathrm{p} \times \mathrm{q}$ and $\mathrm{B}$ is of order $\mathrm{q} \times \mathrm{r}$ what is the order of $\mathrm{AB}$ and $\mathrm{BA}$ ?
Solution:
If $A$ is of order $p \times q[\because p \times q q \times r=p \times r]$
the order of $A B=p \times r[\because q \times r p \times q=r \neq p]$
Product of BA cannot be defined/found as the number of columns in $\mathrm{B} \neq$. The number of rows in A.
 

Question 2.
If $A$ is of order $p \times q$ and $B$ is of order $q \times r$ what is the order of $A B$ and $B A$ ?
Answer:
Order of $\mathrm{A}=\mathrm{a} \times(\mathrm{a}+3)$
Order of $B=b \times(17-b)$
Given: Product of $\mathrm{AB}$ exist
$a+3=b$
$a-b=-3 \ldots .(1)$
Product of BA exist
$17-b=a$
$-a-b=-17$
$a+b=17$
(2)
$(1)+(2) \Rightarrow 2 a=14$
$a=\frac{14}{2}=7$
Substitute the value of $a=7$ in ( 1 )
$7-\mathrm{b}=-3 \Rightarrow-\mathrm{b}=-3-7$
$-\mathrm{b}=-10 \Rightarrow \mathrm{b}=10$
The value of $\mathrm{b}=7$ and $\mathrm{b}=10$
 

Question $3 .$
Find the order of the product matrix $\mathrm{AB}$ if

Solution:

Question $4 .$
If $\mathrm{A}=\left[\begin{array}{ll}2 & 5 \\ 4 & 3\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}1 & -3 \\ 2 & 5\end{array}\right]$ find $\mathrm{AB}, \mathrm{BA}$ and check if $\mathrm{AB}=\mathrm{BA}$ ?
Solution:
$\mathrm{AB}=\left[\begin{array}{ll}2 & 5 \\ 4 & 3\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ 2 & 5\end{array}\right]$
$\begin{aligned}
& =\left[\begin{array}{ll}
\left(\begin{array}{ll}
2 & 5
\end{array}\right)\left(\begin{array}{l}
1 \\
2
\end{array}\right) & \left(\begin{array}{ll}
2 & 5
\end{array}\right)\left(\begin{array}{l}
-3 \\
5
\end{array}\right) \\
\left(\begin{array}{ll}
4 & 3
\end{array}\right)\left(\begin{array}{l}
1 \\
2
\end{array}\right) & \left(\begin{array}{ll}
4 & 3
\end{array}\right)\left(\begin{array}{l}
-3 \\
5
\end{array}\right)
\end{array}\right] \\
& =\left[\begin{array}{ll}
(2+10) & (-6+25) \\
(4+6) & (-12+15)
\end{array}\right]=\left[\begin{array}{cc}
12 & 19 \\
10 & 3
\end{array}\right] \ldots(1)
\end{aligned}$

$\mathrm{BA}=\left[\begin{array}{cc}
1 & -3 \\
2 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 5 \\
4 & 3
\end{array}\right]$

$\begin{aligned}
& =\left[\begin{array}{cc}
(2-12) & (5-9) \\
(4+20) & (10+15)
\end{array}\right]=\left[\begin{array}{cc}
-10 & -4 \\
24 & 25
\end{array}\right] \\
& \text { (1) } \neq(2) \therefore \mathrm{AB} \neq \mathrm{BA} \\
&
\end{aligned}$
Question $5 .$
Given that $A=\left[\begin{array}{cc}1 & 3 \\ 5 & -1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 5 & 2\end{array}\right]$, $C=\left[\begin{array}{ccc}1 & 3 & 2 \\ -4 & 1 & 3\end{array}\right]$ verify that $A(B+C)=$
$\mathbf{A B}+\mathbf{A C}$.
Solution:
$\begin{aligned}
&\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC} \text {. }\\
&\text { L.H.S }=\mathrm{A}(\mathrm{B}+\mathrm{C})\\
&(B+C)=\left(\begin{array}{ccc}
1 & -1 & 2 \\
3 & 5 & 2
\end{array}\right)+\left(\begin{array}{ccc}
1 & 3 & 2 \\
-4 & 1 & 3
\end{array}\right)\\
&=\left(\begin{array}{ccc}
2 & 2 & 4 \\
-1 & 6 & 5
\end{array}\right)\\
&A(B+C)=\left(\begin{array}{cc}
1 & 3 \\
5 & -1
\end{array}\right)\left(\begin{array}{ccc}
2 & 2 & 4 \\
-1 & 6 & 5
\end{array}\right)\\
&=\left(\begin{array}{lll}
\left(\begin{array}{ll}
1 & 3
\end{array}\right)\left(\begin{array}{l}
2 \\
-1
\end{array}\right) & \left(\begin{array}{ll}
1 & 3
\end{array}\right)\left(\begin{array}{l}
2 \\
6
\end{array}\right) & \left(\begin{array}{ll}
1 & 3
\end{array}\right)\left(\begin{array}{l}
4 \\
5
\end{array}\right) \\
\left(\begin{array}{ll}
5 & -1
\end{array}\right)\left(\begin{array}{l}
2 \\
-1
\end{array}\right) & \left(\begin{array}{ll}
5 & -1
\end{array}\right)\left(\begin{array}{l}
2 \\
6
\end{array}\right) & \left(\begin{array}{ll}
5 & -1
\end{array}\right)\left(\begin{array}{l}
4 \\
5
\end{array}\right)
\end{array}\right)\\
&=\left[\begin{array}{lll}
(2-3) & (2+18) & (4+15) \\
(10+1) & (10-6) & (20-5)
\end{array}\right]\\
&=\left[\begin{array}{ccc}
-1 & 20 & 19 \\
11 & 4 & 15
\end{array}\right]\\
&\mathrm{AB}=\left[\begin{array}{cc}
1 & 3 \\
5 & -1
\end{array}\right]\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 5 & 2
\end{array}\right]
\end{aligned}$

$=\left[\begin{array}{ccc}(1+9) & (-1+15) & (2+6) \\ (5-3) & (-5-5) & (10-2)\end{array}\right]$
$=\left[\begin{array}{ccc}10 & 14 & 8 \\ 2 & -10 & 8\end{array}\right]$
$A C=\left[\begin{array}{cc}1 & 3 \\ 5 & -1\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ -4 & 1 & 3\end{array}\right]$
$=\left[\begin{array}{lll}(1-12) & (3+3) & (2+9) \\ (5+4) & (15-1) & (10-3)\end{array}\right]$
$=\left[\begin{array}{ccc}-11 & 6 & 11 \\ 9 & 14 & 7\end{array}\right]$
$\mathrm{AB}+\mathrm{AC}=\left[\begin{array}{ccc}10 & 14 & 8 \\ 2 & -10 & 8\end{array}\right]+\left[\begin{array}{ccc}-11 & 6 & 11 \\ 9 & 14 & 7\end{array}\right]$
$=\left[\begin{array}{ccc}-1 & 20 & 19 \\ 11 & +4 & 15\end{array}\right]$
$(1)=(2) \Rightarrow$ L.H.S. $=$ R.H.S.
$\therefore \mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC}$ verified.

Question $6 .$
Show that the matrices $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}1 & -2 \\ -3 & 1\end{array}\right]$ satisfy commutative property $\mathrm{AB}=\mathrm{BA}$
Solution:
$\begin{aligned}
\mathrm{A} &=\left[\begin{array}{ll}
1 & 2 \\
3 & 1
\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}
1 & -2 \\
-3 & 1
\end{array}\right] \\
\mathrm{AB} &=\left[\begin{array}{ll}
1 & 2 \\
3 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
-3 & 1
\end{array}\right] \\
&=\left[\begin{array}{ll}
(1-6) & (-2+2) \\
(3-3) & (-6+1)
\end{array}\right]=\left[\begin{array}{cc}
-5 & 0 \\
0 & -5
\end{array}\right] \ldots(1) \\
\mathrm{BA} &=\left[\begin{array}{cc}
1 & -2 \\
-3 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & 2 \\
3 & 1
\end{array}\right] \\
&=\left[\begin{array}{cc}
(1-6) & (2-2) \\
(-3+3) & (-6+1)
\end{array}\right]=\left[\begin{array}{cc}
-5 & 0 \\
0 & -5
\end{array}\right]
\end{aligned}$

$\therefore(1)=(2)$
$\therefore \mathrm{AB}=\mathrm{BA}$ verified.

Question 7.
Let $A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right], C=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$ show that
(i) $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$
(ii) $(A-B) C=(A C-B C)$
(iii) $(A-B)^{T}=A^{T}-B^{T}$
Solution:
(i) $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$

$\begin{aligned}
&\text { L.H.S. }=\mathrm{A}(\mathrm{BC})\\
&(B C)=\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
(8+0) & (0+0) \\
(2+5) & (0+10)
\end{array}\right]\\
&=\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right]\\
&\mathrm{A}(\mathrm{BC})=\left[\begin{array}{ll}
1 & 2 \\
1 & 3
\end{array}\right]\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right]=\left[\begin{array}{ll}
(8+14) & (0+20) \\
(8+21) & (0+30)
\end{array}\right]\\
&=\left[\begin{array}{ll}
22 & 20 \\
29 & 30
\end{array}\right]\\
&\text { R.H.S }=(\mathrm{AB}) \mathrm{C}\\
&A B=\left(\begin{array}{ll}
1 & 2 \\
1 & 3
\end{array}\right)\left(\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right)=\left[\begin{array}{ll}
(4+2) & (0+10) \\
(4+3) & (0+15)
\end{array}\right]\\
&=\left[\begin{array}{ll}
6 & 10 \\
7 & 15
\end{array}\right]\\
&(\mathrm{AB}) \mathrm{C}=\left[\begin{array}{ll}
6 & 10 \\
7 & 15
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
(12+10) & (0+20) \\
(14+15) & (0+30)
\end{array}\right]\\
&=\left[\begin{array}{ll}
22 & 20 \\
29 & 30
\end{array}\right]
\end{aligned}$

$(1)=(2) \Rightarrow$ L.H.S. $=$ R.H.S.
$\therefore A(B C)=(A B) C$, verified.
(ii) $(\mathbf{A}-\mathbf{B}) \mathbf{C}=\mathbf{A C}-\mathbf{B C}$
L.H.S. $=(A-B) C$
$A-B=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}-3 & 2 \\ 0 & -2\end{array}\right]$
$(A-B) C=\left[\begin{array}{cc}-3 & 2 \\ 0 & -2\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{cc}(-6+2) & (0+4) \\ (0-2) & (0-4)\end{array}\right]$
$=\left[\begin{array}{cc}-4 & 4 \\ -2 & -4\end{array}\right]$
R.H.S $=\mathrm{AC}-\mathrm{BC}$
$\mathrm{AC}=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}(2+2) & (0+4) \\ (2+3) & (0+6)\end{array}\right]$
$=\left[\begin{array}{ll}4 & 4 \\ 5 & 6\end{array}\right]$
$\mathrm{BC}=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{ll}(8+0) & (0+0) \\ (2+5) & (0+10)\end{array}\right]$

$\begin{aligned}
&=\left[\begin{array}{ll}
(8+0) & (0+0) \\
(2+5) & (0+10)
\end{array}\right] \\
&=\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right] \\
A C &-B C=\left[\begin{array}{ll}
4 & 4 \\
5 & 6
\end{array}\right]-\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right]=\left[\begin{array}{cc}
-4 & 4 \\
-2 & -4
\end{array}\right]
\end{aligned}$
$(1)=(2) \Rightarrow$ LHS $=$ RHS. Hence verified.
(iii)
$\begin{aligned}
(A-B)^{\mathrm{T}}=A^{\mathrm{T}}-\mathbf{B}^{\mathrm{T}} \\
\text { L.H.S }=(\mathrm{A}-\mathrm{B}) &=\left[\begin{array}{ll}
1 & 2 \\
1 & 3
\end{array}\right]-\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right] \\
&=\left[\begin{array}{cc}
-3 & 2 \\
0 & -2
\end{array}\right] \\
\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right], \mathrm{B}^{\mathrm{T}}=\left[\begin{array}{cc}
-3 & 0 \\
2 & -2
\end{array}\right]
\end{aligned}$

$\begin{aligned}
A^{\top}=\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right], \mathrm{B}^{\top} &=\left[\begin{array}{ll}
4 & 1 \\
0 & 5
\end{array}\right] \\
\mathrm{A}^{\top}-\mathrm{B}^{\top} &=\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right]-\left[\begin{array}{ll}
4 & 1 \\
0 & 5
\end{array}\right] \\
&=\left[\begin{array}{cc}
-3 & 0 \\
2 & -2
\end{array}\right]
\end{aligned}$
$(1)=(2) .$ L.H.S. = R.H.S. Hence verified.

Question $8 .$
If $A=\left[\begin{array}{cc}\cos \theta & 0 \\ 0 & \cos \theta\end{array}\right], B=\left[\begin{array}{cc}\sin \theta & 0 \\ 0 & \sin \theta\end{array}\right]$ then show that $\mathbf{A}^{2}+\mathbf{B}^{2}=\mathbf{I}$.
Solution:

L.H.S $=\mathrm{A}^{2}+\mathrm{B}^{2}$
$A^{2}=\left[\begin{array}{cc}\cos \theta & 0 \\ 0 & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & 0 \\ 0 & \cos \theta\end{array}\right]$
$=\left[\begin{array}{cc}\cos ^{2} \theta & 0 \\ 0 & \cos ^{2} \theta\end{array}\right]$
$\mathrm{B}^{2}=\left(\begin{array}{cc}\sin \theta & 0 \\ 0 & \sin \theta\end{array}\right)\left(\begin{array}{cc}\sin \theta & 0 \\ 0 & \sin \theta\end{array}\right)$
$=\left(\begin{array}{cc}\sin ^{2} \theta & 0 \\ 0 & \sin ^{2} \theta\end{array}\right)$
$A^{2}+B^{2}=\left(\begin{array}{cc}\cos ^{2} \theta & 0 \\ 0 & \cos ^{2} \theta\end{array}\right)+\left(\begin{array}{cc}\sin ^{2} \theta & 0 \\ 0 & \sin ^{2} \theta\end{array}\right)$
$=\left(\begin{array}{cc}\sin ^{2} \theta+\cos ^{2} \theta & 0 \\ 0 & \sin ^{2} \theta+\cos ^{2} \theta\end{array}\right)$
$=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=I=$ R.H.S.
Hence proved.
 

Question 9.
If $A=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right)$ prove that $A A^{T}=I$

Solution:
$\begin{gathered}
\mathrm{A}^{T}=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right) \\
\mathrm{A} \cdot \mathrm{A}^{\top}=\left(\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right)\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right) \\
=\left(\begin{array}{cc}
\cos ^{2} \theta+\sin ^{2} \theta & -\cos \theta \sin \theta+\cos ^{2} \theta \sin \theta \\
-\sin \theta \cos \theta+\cos ^{2} \sin \theta & \sin ^{2} \theta+\cos ^{2} \theta
\end{array}\right) \\
=\left(\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right)=I
\end{gathered}$
Hence it is proved.
 

Question $10 .$
Verify that $\mathrm{A}^{2}=\mathrm{I}$ when $\mathrm{A}=\left(\begin{array}{ll}5 & -4 \\ 6 & -5\end{array}\right)$
Solution:
$\begin{aligned}
A &=\left(\begin{array}{ll}
5 & -4 \\
6 & -5
\end{array}\right) \\
A^{2} &=\left(\begin{array}{ll}
5 & -4 \\
6 & -5
\end{array}\right)\left(\begin{array}{ll}
5 & -4 \\
6 & -5
\end{array}\right) \\
&=\left(\begin{array}{ll}
(25-24) & (-20+20) \\
(30-30) & (-24+25)
\end{array}\right) \\
&=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\mathrm{I} .
\end{aligned}$
Hence it is proved.

Question $11 .$
If $A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$ and $I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ show that $\mathbf{A}^{2}-(a+d) \mathbf{A}=(b c-a d) I_{2^{*}}$
Solution:
L.H.S $=\mathrm{A}^{2}-(a+d) \mathrm{A}$
$\mathrm{A}^{2}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}a^{2}+b c & a d+b d \\ a c+c d & b c+d^{2}\end{array}\right)$
$(a+d) \mathrm{A}=(a+d)\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$
$=\left[\begin{array}{ll}a^{2}+a d & a b+b d \\ a c+c d & a d+d^{2}\end{array}\right]$
$\mathrm{A}^{2}-(a+d) \mathrm{A}$
$=\left[\begin{array}{ll}a^{2}+b c & a d+b d \\ a c+c d & b c+d^{2}\end{array}\right]-\left[\begin{array}{ll}a^{2}+a d & a b+b d \\ a c+c d & a d+d^{2}\end{array}\right]$
$=\left[\begin{array}{cc}b c-a d & 0 \\ 0 & b c-a d\end{array}\right]$
$=(b c-a d)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=(b c-a d) \mathrm{I}_{2}=$ R.H.S.
Hence it is proved.

Question $12 .$
If $A=\left[\begin{array}{lll}5 & 2 & 9 \\ 1 & 2 & 8\end{array}\right], B=\left[\begin{array}{cc}1 & 7 \\ 1 & 2 \\ 5 & -1\end{array}\right]$ verify that $(\mathbf{A B})^{\mathrm{T}}=\mathbf{B}^{\top} \mathbf{A}^{\mathrm{T}}$.
Solution:
$\mathrm{A}=\left[\begin{array}{lll}
5 & 2 & 9 \\
1 & 2 & 8
\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}
1 & 7 \\
1 & 2 \\
5 & -1
\end{array}\right]$
S.T. $(A B)^{\mathrm{T}}=B^{\top} A^{T}$
L.H.S $=(A B)^{T}$
$\begin{aligned}
(\mathrm{AB}) &=\left[\begin{array}{lll}
5 & 2 & 9 \\
1 & 2 & 8
\end{array}\right]\left[\begin{array}{cc}
1 & 7 \\
1 & 2 \\
5 & -1
\end{array}\right] \\
&=\left[\begin{array}{cc}
(5+2+45) & (35+4-9) \\
(1+2+40) & (7+4-8)
\end{array}\right]=\left[\begin{array}{cc}
52 & 30 \\
43 & 3
\end{array}\right]
\end{aligned}$
$(A B)^{\mathrm{T}}=\left[\begin{array}{cc}52 & 43 \\ 30 & 3\end{array}\right]$

$\begin{aligned}
\mathrm{B}^{\mathrm{T}} &=\left[\begin{array}{ccc}
1 & 1 & 5 \\
7 & 2 & -1
\end{array}\right], \mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ll}
5 & 1 \\
2 & 2 \\
9 & 8
\end{array}\right] \\
\mathrm{B}^{\top} \mathrm{A}^{\top} &=\left[\begin{array}{cc}
(5+2+45) & (1+2+40) \\
(35+4-9) & (7+4-8)
\end{array}\right] \\
&=\left[\begin{array}{cc}
52 & 43 \\
30 & 3
\end{array}\right]
\end{aligned}$
$(1)=(2) \Rightarrow$ L.H.S. $=$ R.H.S. Verified.
 

Question 13.
If $\mathrm{A}=$ show that $\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}_{2}=0$.
Solution:
L.H.S $=\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}_{2}$
$A^{2}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}(9-1) & (3+2) \\ (-3-2) & (-1+4)\end{array}\right]$

$\begin{aligned}
&=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right] \\
5 \mathrm{~A} &=5\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right] \\
7 \mathrm{I}_{2}^{\prime} &=\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] \\
\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}_{2} &=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] \\
&=\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]=0
\end{aligned}$
Hence verified.