Question 13 Marks
If $A=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$, then show that $A^2-4 A+3l=0$.
Answer$
\begin{aligned}
A ^2 & = A \cdot A =\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right] \\
& =\left[\begin{array}{rr}
4+1 & -2-2 \\
-2-2 & 1+4
\end{array}\right]=\left[\begin{array}{rr}
5 & -4 \\
-4 & 5
\end{array}\right] \\
\therefore & A ^2-4 A +3 I \\
& =\left[\begin{array}{rr}
5 & -4 \\
-4 & 5
\end{array}\right]-4\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]+3\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{aligned}
$
$
\begin{aligned}
& =\left[\begin{array}{rr}
5 & -4 \\
-4 & 5
\end{array}\right]-\left[\begin{array}{rr}
8 & -4 \\
-4 & 8
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& =\left[\begin{array}{rr}
5-8+3 & -4-(-4)+0 \\
-4-(-4)+0 & 5-8+3
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
\therefore & A^2-4 A+3 I=0 .
\end{aligned}
$
View full question & answer→Question 23 Marks
If $A=\left[\begin{array}{ll}3 & 1 \\ 1 & 5\end{array}\right], B=\left[\begin{array}{cc}1 & 2 \\ 5 & -2\end{array}\right]$, verify $|A B|=|A||B|$.
Answer$
\begin{aligned}
& AB =\left[\begin{array}{ll}
3 & 1 \\
1 & 5
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
5 & -2
\end{array}\right] \\
&=\left[\begin{array}{rr}
3+5 & 6-2 \\
1+25 & 2-10
\end{array}\right]=\left[\begin{array}{rr}
8 & 4 \\
26 & -8
\end{array}\right] \\
& \therefore| AB |=\left|\begin{array}{rr}
8 & 4 \\
26 & -8
\end{array}\right| \\
&=-64-104=-168.....(1) \\
&| A |=\left|\begin{array}{rr}
3 & 1 \\
1 & 5
\end{array}\right|=15-1=14 \\
&| B |=\left|\begin{array}{rr}
1 & 2 \\
5 & -2
\end{array}\right|=-2-10=-12 \\
& \therefore| A || B |=14(-12)=-168....(2)
\end{aligned}
$
From (1) and (2), $| AB |=| A || B |$.
View full question & answer→Question 33 Marks
Solve the following equations by method of reduction $: 2x + y = 5, 3x – 5y = -3$
AnswerThe given equation can be written in matrix form as:
$ \left[\begin{array}{ll} 2 & 1 \\ 3 & 5 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 5 \\ -3 \end{array}\right]$
By $R_2-5 R_1$, we get
${\left[\begin{array}{rr} 2 & 1 \\ -7 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 5 \\ -28 \end{array}\right]}$
$\therefore\left[\begin{array}{r} 2 x+y \\ -7 x+0 \end{array}\right]=\left[\begin{array}{r} 5 \\ -28 \end{array}\right]$
By equality of matrices,
$ 2 x+y=5$
$ -7 x=-28 $
From$ (2), x=4$
Substituting $x=4$ in $(1),$ we get
$ 2(4)+y=5$
$ \therefore y=-3 $
Hence, $x=4$ and $y=-3$ is the required solution.
View full question & answer→Question 43 Marks
Find inverse of the following matrices (if they exist) by elementary transformation : $\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$
AnswerLet $A =\left(\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right)$
Then $|A|=\left|\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right|=8-7=1 \neq 0$
$\therefore A ^{-1}$ exist.
We write $A ^{-1} A = I$
$
\therefore A^{-1}\left(\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right)=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)
$
By $C_1-C_2$, we get
$
A^{-1}\left[\begin{array}{ll}
1 & 1 \\
3 & 4
\end{array}\right]=\left(\begin{array}{rr}
1 & 0 \\
-1 & 1
\end{array}\right)
$
By $C_2-C_1$, we get
$
A^{-1}\left(\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]=\left[\begin{array}{rr}
1 & -1 \\
-1 & 2
\end{array}\right]
$
By $C_1-3 C_2$, we get
$
\begin{aligned}
& A^{-1}\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left[\begin{array}{rr}
4 & -1 \\
-7 & 2
\end{array}\right) \\
& \therefore A^{-1}=\left(\begin{array}{rr}
4 & -1 \\
-7 & 2
\end{array}\right) .
\end{aligned}
$
View full question & answer→Question 53 Marks
Find inverse of the following matrices (if they exist) by elementary transformation : $\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
AnswerLet $A=\left(\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right)$
Then $|A|=\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|=3-(-2)=5 \neq 0$
$\therefore A ^{-1}$ exists.
We write, $AA ^{-1}= I$
$
\therefore\left(\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right) A ^{-1}=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)
$
By $R_2-2 R_1$, we get
$
\left(\begin{array}{cc}
1 & -1 \\
0 & 5
\end{array}\right] A^{-1}=\left(\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right)
$
By $\left(\frac{1}{5}\right) R_2$, we get
$
\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right] A^{-1}=\left[\begin{array}{rr}
1 & 0 \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]
$
By $R_1+R_2$, we get
$
\begin{aligned}
& {\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] A ^{-1}=\left[\begin{array}{rr}
3 & \frac{1}{5} \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]} \\
& \therefore A ^{-1}=\left[\begin{array}{rr}
\frac{3}{5} & \frac{1}{5} \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]
\end{aligned}
$
View full question & answer→Question 63 Marks
if $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right]$, then reduce it to unit matrix by row tranformation.
Answer$
\begin{aligned}
& A B=\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right] \\
& \therefore|A|=\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right] \\
& =1(1-0)-0(2-0)+0(6-3) \\
& =1-0+0 \\
& =1 \neq 0
\end{aligned}
$
$\therefore A$ is non-singular matrix.
Hence, row transformations are possible.
Now, $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right]$
Applying $R_2 \rightarrow R_2-2 R_1$ and $R_3 \rightarrow R_3-3 R_1$, we get
$
A=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 3 & 1
\end{array}\right]
$
Applying $R_3 \rightarrow R_3-3 R_2$, we get
$A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
View full question & answer→Question 73 Marks
Express the following equations in matrix form and solve em by method of reduction $: x + y + z = 1, 2x + 3y + 2z = 2$ and $x + y + 2z = 4.$
AnswerThe given equation can be written in matrix form as:
$ \left[\begin{array}{ll} 2 & 1 \\ 3 & 5 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 5 \\ -3
\end{array}\right]$
By $R_2-5 R_1$, we get
${\left[\begin{array}{rr} 2 & 1 \\ -7 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 5 \\ -28 \end{array}\right]}$
$ \therefore\left[\begin{array}{r} 2 x+y \\ -7 x+0 \end{array}\right]=\left[\begin{array}{r} 5 \\ -28
\end{array}\right]$
By equality of matrices,
$ 2 x+y=5$
$ -7 x=-28 $
From$ (2), x=4$
Substituting $x=4$ in $(1),$ we get
$ 2(4)+y=5$
$ \therefore y=-3 $
Hence, $x=4$ and $y=-3$ is the required solution.
View full question & answer→Question 83 Marks
Express the following equations in matrix form and solve em by method of reduction : $3x – y = 1, 4x + y = 6.$
AnswerThe given equations can be written in the matrix form as:
$\left[\begin{array}{rr}3 & -1 \\4 & 1\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{l}1 \\6\end{array}\right]$
By $4 R_1$ and $3 R_2$, we get
$\left[\begin{array}{rr}12 & -4 \\12 & 3\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{r}4 \\18\end{array}\right]$
By $R_2-R_1$, we get
$\begin{aligned}& \left[\begin{array}{rr}12 & -4 \\0 & 7\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{r}4 \\14\end{array}\right]\end{aligned} $
$\therefore \left[\begin{array}{r}12 x-4 y \\0+7 y\end{array}\right]=\left(\begin{array}{r}4 \\14 \end{array}\right)$
By equality of matrices,
$\begin{aligned}& 12 x-4 y=4 . . . .(1) \\& 7 y=14 . . . .(2)\end{aligned}$
From $(2), y=2$
Substituting $y=2$ in $(1),$ we get
$12 x-8=4$
$ \therefore 12 x=12$
$ \therefore x=1$
Hence, $x=1, y=2$ is the required solution.
View full question & answer→Question 93 Marks
Express the following equations in matrix form and solve em by method of reduction : $x + 3y = 2, 3x + 5y = 4$.
AnswerThe given equations can be written in the matrix form as:
$ \left[\begin{array}{ll} 1 & 3 \\ 3 & 5 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 2 \\ 4 \end{array}\right]$
By $R_2-3 R_1$, we get
$ \left(\begin{array}{rr} 1 & 3 \\ 0 & -4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 2 \\ -2 \end{array}\right]$
$\therefore \left(\begin{array}{l} x+3 y \\ 0-4 y \end{array}\right]=\left[\begin{array}{r} 2 \\ -2 \end{array}\right]$
By equality of matrices,
$ \begin{aligned} & x+3 y=2 \\ & -4 y=-2 \end{aligned}$
From $ (2), y=\frac{1}{2}$
Substituting $y=\frac{1}{2}$ in $(1),$ we get
$ \begin{aligned} & x+\frac{3}{2}=2 \\ \therefore & x=2-\frac{3}{2}=\frac{1}{2} \end{aligned}$
Hence, $x=\frac{1}{2}, y=\frac{1}{2}$ is the required solution.
View full question & answer→Question 103 Marks
Find the inverses of the following matrices by the adjoint mathod : $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
AnswerLet $A =\left(\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right)$
Then $|A|=\left|\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right|$
$
\begin{aligned}
& =1(10-0)-2(0-0)+3(0-0) \\
& =10 \neq 0
\end{aligned}
$
$\therefore A ^{-1}$ exist. First we have to find the cofactor matrix $=\left[ A _{i j}\right]_{3 \times 3}$, where $A _{i j}=(-1)^{i+j} M _{i j}$
Now, $A_{11}=(-1)^{1+1} M_{11}=\left|\begin{array}{ll}2 & 4 \\ 0 & 5\end{array}\right|=10-0=10$
$
\begin{aligned}
& A_{12}=(-1)^{1+2} M_{12}=-\left|\begin{array}{ll}
0 & 4 \\
0 & 5
\end{array}\right|=-(0-0)=0 \\
& A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right|=0-0=0 \\
& A_{21}=(-1)^{2+1} M_{21}=-\left|\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right|=-(10-0)=-10
\end{aligned}
$
$
\begin{aligned}
& A_{22}=(-1)^{2+2} M_{22}=\left|\begin{array}{ll}
1 & 3 \\
0 & 5
\end{array}\right|=5-0=5 \\
& A_{23}=(-1)^{2+3} M_{23}=-\left|\begin{array}{ll}
1 & 2 \\
0 & 0
\end{array}\right|=-(0-0)=0 \\
& A_{31}=(-1)^{3+1} M_{31}=\left|\begin{array}{ll}
2 & 3 \\
2 & 4
\end{array}\right|=8-6=2 \\
& A_{32}=(-1)^{3+2} M_{32}=-\left|\begin{array}{ll}
1 & 3 \\
0 & 4
\end{array}\right|=-(4-0)=-4 \\
& A_{33}=(-1)^{3+3} M_{33}=\left|\begin{array}{ll}
1 & 2 \\
0 & 2
\end{array}\right|=2-0=2
\end{aligned}
$
$\begin{aligned} & \therefore \text { the cofactor matrix } \\ & =\left(\begin{array}{lll} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33}\end{array}\right)=\left[\begin{array}{rrr}10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2\end{array}\right] \\ & \therefore \text { adj } A =\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right] \\ & \therefore A ^{-1}=\frac{1}{| A |} \text { (adj A) } \\ & \therefore A ^{-1}=\frac{1}{10}\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]\end{aligned}$
View full question & answer→Question 113 Marks
Find the inverses of the following matrices by the adjoint mathod : $\left[\begin{array}{cc}2 & -2 \\ 4 & 5\end{array}\right]$
Answer$
\begin{aligned}
& \text { Let } A=\left[\begin{array}{cc}
2 & -2 \\
4 & 5
\end{array}\right] \\
& \therefore|A|=\left[\begin{array}{cc}
2 & -2 \\
4 & 5
\end{array}\right]=10+8=18 \neq 0 \\
& \therefore A^{-1} \text { exists. } \\
& A_{11}=(-1)^{1+1} M_{11}=(1)(5)=5 \\
& A_{12}=(-1)^{1+2} M_{12}=(-1)(4)=-4 \\
& A_{21}=(-1)^{2+1} M_{21}=(-1)(-2)=2 \\
& A_{22}=(-1)^{2+2} M_{22}=(1)(2)=2
\end{aligned}
$
$\therefore$ The matrix of the co-factors is
$
\begin{aligned}
& {\left[ A _{ ij }\right]_{2 \times 2}=\left[\begin{array}{ll}
A _{11} & A _{12} \\
A _{21} & A _{22}
\end{array}\right]=\left[\begin{array}{cc}
5 & -4 \\
2 & 2
\end{array}\right]} \\
& \text { Now } \operatorname{adj} A=\left[ A _{ ij }\right]_{2 \times 2}^{\top}=\left[\begin{array}{cc}
5 & 2 \\
-4 & 2
\end{array}\right] \\
& \therefore A -1=\frac{1}{| A |}(\operatorname{adj} A ) \\
& =\frac{1}{18}\left[\begin{array}{cc}
5 & 2 \\
-4 & 2
\end{array}\right] .
\end{aligned}
$
View full question & answer→Question 123 Marks
Find the inverses of the following matrices by the adjoint mathod : $\left[\begin{array}{ll}3 & -1 \\ 2 & -1\end{array}\right]$
AnswerLet $A =\left(\begin{array}{ll}3 & -1 \\ 2 & -1\end{array}\right)$
Then $|A|=\left|\begin{array}{ll}3 & -1 \\ 2 & -1\end{array}\right|=-3-(-2)=-1 \neq 0$
$\therefore A ^{-1}$ exists.
First we have to find the cofactor matrix $=\left[ A _{i j}\right]_{2 \times 2}$, where $A _{i j}=(-1)^{i+j} M _{i j}$
Now, $A_{11}=(-1)^{1+1} M_{11}=-1$
$
\begin{aligned}
& A_{12}=(-1)^{1+2} M_{12}=-2 \\
& A_{21}=(-1)^{2+1} M_{21}=-(-1)=1 \\
& A_{22}=(-1)^{2+2} M_{22}=3
\end{aligned}
$
$\therefore$ the cofactor matrix
$
\begin{aligned}
& =\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]=\left[\begin{array}{rr}
-1 & -2 \\
1 & 3
\end{array}\right] \\
& \therefore \operatorname{adj} A=\left[\begin{array}{ll}
-1 & 1 \\
-2 & 3
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-1}\left[\begin{array}{ll}
-1 & 1 \\
-2 & 3
\end{array}\right] \\
& \therefore A^{-1}=\left[\begin{array}{ll}
1 & -1 \\
2 & -3
\end{array}\right] .
\end{aligned}
$
View full question & answer→Question 133 Marks
Transform $\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4\end{array}\right]$ into an upper triangularmatrix by suitable row transformations.
AnswerLet $A=\left(\begin{array}{rrr}1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4\end{array}\right)$
By $R_2-2 R_1$ and $R_3-3 R_1$, we get
$
A \sim\left(\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 5 & -2
\end{array}\right)
$
By $\left(\frac{1}{3}\right) R_2$, we get
$
A \sim\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 1 & -\frac{1}{3} \\
0 & 5 & -2
\end{array}\right]
$
By $R_3-5 R_2$, we get
$
A \sim\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 1 & -\frac{1}{3} \\
0 & 0 & -\frac{1}{3}
\end{array}\right]
$
This is an upper triangular matrix.
View full question & answer→Question 143 Marks
Find matrix $X,$ if $A X=B$, where $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
AnswerThe given equation can be written in matrix form as:
$ \left[\begin{array}{ll} 2 & 1 \\ 3 & 5 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 5 \\ -3
\end{array}\right]$
By $R_2-5 R_1$, we get
$\begin{aligned} & {\left[\begin{array}{rr} 2 & 1 \\ -7 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} 5 \\ -28 \end{array}\right]} \\ & \therefore\left[\begin{array}{r} 2 x+y \\ -7 x+0 \end{array}\right]=\left(\begin{array}{r} 5 \\ -28
\end{array}\right] \end{aligned}$
By equality of matrices,
$ 2 x+y=5$
$ -7 x=-28 $
From$ (2), x=4$
Substituting $x=4$ in $(1),$ we get
$ 2(4)+y=5$
$ \therefore y=-3 $
Hence, $x=4$ and $y=-3$ is the required solution.
View full question & answer→Question 153 Marks
If $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=2(i-j)$. Find $A$ and $A^{\top}$. State whether $A$ and $A^{\top}$ both are symmetric or skew-symmetric matrices.
Answer$
A =\left[a_{i j}\right]_{3 \times 3}=\left(\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right)
$
Given : $a_{i j}=2(i-j)$
$
\begin{aligned}
\therefore a_{11} & =2(1-1)=0, a_{12}=2(1-2)=-2, \\
a_{13} & =2(1-3)=-4, a_{21}=2(2-1)=2, \\
a_{22} & =2(2-2)=0, a_{23}=2(2-3)=-2, \\
a_{31} & =2(3-1)=4, a_{32}=2(3-2)=2, \\
a_{33} & =2(3-3)=0
\end{aligned}
$
$
\therefore A=\left(\begin{array}{rrr}
0 & -2 & -4 \\
2 & 0 & -2 \\
4 & 2 & 0
\end{array}\right)
$
$
\begin{aligned}
& \therefore A ^{ T }=\left(\begin{array}{rrr}
0 & 2 & 4 \\
-2 & 0 & 2 \\
-4 & -2 & 0
\end{array}\right] \\
& \therefore- A ^{ T }=-\left(\begin{array}{rrr}
0 & 2 & 4 \\
-2 & 0 & 2 \\
-4 & -2 & 0
\end{array}\right)=\left[\begin{array}{rrr}
0 & -2 & -4 \\
2 & 0 & -2 \\
4 & 2 & 0
\end{array}\right) \\
& \therefore A =- A ^{ T } \text { and } A ^{ T }=- A
\end{aligned}
$
Hence, $A$ and $A^{\top}$ are both skew-symmetric matrices.
View full question & answer→Question 163 Marks
If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$, find $k$ so that $A^2-8 A-k l=O$, where $I$ is a $2 \times 2$ unit matrix and $O$ is null matrix of order 2 .
Answer$
\begin{aligned}
A ^2= A \cdot A & =\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right] \\
= & {\left[\begin{array}{rr}
1-0 & 0+0 \\
-1-7 & 0+49
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right] } \\
\therefore A ^2-8 A -k I & =\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]-8\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]-k\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]-\left[\begin{array}{rr}
8 & 0 \\
-8 & 56
\end{array}\right]-\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right] \\
& =\left[\begin{array}{cc}
1-8-k & 0-0-0 \\
-8+8-0 & 49-56-k
\end{array}\right] \\
& =\left[\begin{array}{cc}
-k-7 & 0 \\
0 & -k-7
\end{array}\right]
\end{aligned}
$
But, $A ^2-8 A -k I =0$
$
\therefore\left[\begin{array}{cc}
-k-7 & 0 \\
0 & -k-7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right)
$
By equality of matrices,
$
\begin{aligned}
& - k -7=0 \\
& \therefore k =-7 .
\end{aligned}
$
View full question & answer→Question 173 Marks
If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, show that $A^2-4 A$ is a scalar matrix.
Answer$
\begin{aligned}
& A^2=A \cdot A=\left(\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right)\left(\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right) \\
& =\left(\begin{array}{lll}
1+4+4 & 2+2+4 & 2+4+2 \\
2+2+4 & 4+1+4 & 4+2+2 \\
2+4+2 & 4+2+2 & 4+4+1
\end{array}\right) \\
& =\left(\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right) \\
& \therefore A ^2-4 A =\left(\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right)-4\left(\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right) \\
& =\left(\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right)-\left(\begin{array}{lll}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4
\end{array}\right) \\
& =\left(\begin{array}{lll}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{array}\right) \\
&
\end{aligned}
$
which is a scalar matrix.
View full question & answer→Question 183 Marks
If $A+I=\left[\begin{array}{ccc}1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3\end{array}\right]$, find the product $(A+I)(A-I)$.
Answer$
\begin{aligned}
& A - I =( A + I )-2 I \\
& =\left(\begin{array}{rrr}
1 & 2 & 0 \\
5 & 4 & 2 \\
0 & 7 & -3
\end{array}\right)-2\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) \\
& =\left(\begin{array}{rrr}
1 & 2 & 0 \\
5 & 4 & 2 \\
0 & 7 & -3
\end{array}\right]-\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right) \\
& =\left[\begin{array}{rrr}
-1 & 2 & 0 \\
5 & 2 & 2 \\
0 & 7 & -5
\end{array}\right] \\
& \therefore( A + I )( A - I )=\left(\begin{array}{rrr}
1 & 2 & 0 \\
5 & 4 & 2 \\
0 & 7 & -3
\end{array}\right)\left[\begin{array}{rrr}
-1 & 2 & 0 \\
5 & 2 & 2 \\
0 & 7 & -5
\end{array}\right) \\
& =\left(\begin{array}{rrr}
-1+10+0 & 2+4+0 & 0+4-0 \\
-5+20+0 & 10+8+14 & 0+8-10 \\
0+35-0 & 0+14-21 & 0+14+15
\end{array}\right) \\
& =\left[\begin{array}{rrr}
9 & 6 & 4 \\
15 & 32 & -2 \\
35 & -7 & 29
\end{array}\right] \\
&
\end{aligned}
$
View full question & answer→Question 193 Marks
$
\begin{aligned}
& \text { Show that } A B=B A \text {, where } A=\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right], B= \\
& {\left[\begin{array}{lll}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]}
\end{aligned}
$
Answer$
\begin{aligned}
AB & =\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\left[\begin{array}{lll}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right] \\
& =\left[\begin{array}{rrr}
-2+6-3 & -6+6-0 & 2-3+1 \\
-1+4-3 & -3+4-0 & 1-2+1 \\
-6+18-12 & -18+18-0 & 6-9+4
\end{array}\right] \\
& =\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)\\
B & =\left[\begin{array}{lll}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right] \\
& =\left[\begin{array}{lll}
-2-3+6 & 3+6-9 & -1-3+4 \\
-4-2+6 & 6+4-9 & -2-2+4 \\
-6-0+6 & 9+0-9 & -3-0+4
\end{array}\right] \\
& =\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$
From (1) and (2), $AB = BA$.
View full question & answer→Question 203 Marks
If $A=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & -3 & 1\end{array}\right], B=\left[\begin{array}{lll}2 & 1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1\end{array}\right]$. State whether $A B=B A$ ? Justify your answer.
Answer$
\begin{aligned}
AB & =\left(\begin{array}{rrr}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right)\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right] \\
& =\left(\begin{array}{rrr}
-2+3+1 & -1+0+2 & -4+2+1 \\
4+9+0 & 2+0+0 & 8+6+0 \\
2-9+1 & 1-0+2 & 4-6+1
\end{array}\right] \\
& =\left(\begin{array}{rrr}
2 & 1 & -1 \\
13 & 2 & 14 \\
-6 & 3 & -1
\end{array}\right) \\
BA & =\left(\begin{array}{rrr}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right)\left[\begin{array}{rrr}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right) \\
& =\left[\begin{array}{rrr}
-2+2+4 & 2+3-12 & 2+0+4 \\
-3+0+2 & 3+0-6 & 3+0+2 \\
-1+4+1 & 1+6-3 & 1+0+1
\end{array}\right] \\
& =\left(\begin{array}{rrr}
4 & -7 & 6 \\
-1 & -3 & 5 \\
4 & 4 & 2
\end{array}\right]
\end{aligned}
$
From (1) and (2), $A B \neq B A$.
View full question & answer→Question 213 Marks
Jay and Ram are two friends. Jay wants to buy $4$ pens and $8$ notebooks. Ram wants to buy $5$ pens and $12$ notebooks. The price of one pen and one notebook was $₹ \ 6$ and $₹ \ 10$ respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.
AnswerThe given data can be written in matrix form as$:$
Number of Pens and Notebooks
$ A =\left[\begin{array}{cc} \text { Pens } & \text { Notebooks } \\ 4 & 8 \\ 5 & 12 \end{array}\right] \text { Jay } $
Price in $₹$
$ B=\left[\begin{array}{l} 6 \\ 10 \end{array}\right] \text { Pen } $
For finding the amount each one of them requires to buy the pens and notebook, we require the multiplication of the two matrices $A$ and $B$.
$ \begin{aligned} \text { Consider } AB & =\left[\begin{array}{rr} 4 & 8 \\ 5 & 12 \end{array}\right]\left[\begin{array}{r} 6 \\ 10 \end{array}\right] \end{aligned}$
$ =\left[\begin{array}{r} 24+80 \\ 30+120 \end{array}\right]$
$=\left(\begin{array}{l} 104 \\ 150 \end{array}\right] $
Hence, Jay requires $₹ \ 104$ and Ram requires $₹ \ 150$ to buy the pens and notebooks.
View full question & answer→Question 223 Marks
Find $x, y, z,$ if $\left\{3\left[\begin{array}{ll}2 & 0 \\ 0 & 2 \\ 2 & 2\end{array}\right]-4\left[\begin{array}{cc}1 & 1 \\ -1 & 2 \\ 3 & 1\end{array}\right]\right\}\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
View full question & answer→Question 233 Marks
$\left\{4\left[\begin{array}{ccc}2 & -1 & 3 \\1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -3 & 4 \\2 & 1 & 1\end{array}\right]\right\}\left[\begin{array}{c}2 \\-1 \\1\end{array}\right]=\left[\begin{array}{l}x \\y\end{array}\right]$
Answer$\left\{4\left[\begin{array}{rrr}2 & -1 & 3 \\ 1 & 0 & 2\end{array}\right]-\left\{\begin{array}{rrr}3 & -3 & 4 \\ 2 & 1 & 1\end{array}\right]\right\}\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\therefore\left\{\left[\begin{array}{rrr}8 & -4 & 12 \\ 4 & 0 & 8\end{array}\right]-\left[\begin{array}{rrr}3 & -3 & 4 \\ 2 & 1 & 1\end{array}\right]\right\}$
$\therefore\left[\begin{array}{lll}5 & -1 & 8 \\ 2 & -1 & 7\end{array}\right]\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\begin{array}{l}\therefore\left[\begin{array}{r}10+1+8 \\ 4+1+7\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]\end{array} $
$\therefore\left[\begin{array}{l}19 \\ 12\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
By equality of matrices, $x=19$ and $y=12$.
View full question & answer→Question 243 Marks
If $A=\left[\begin{array}{cc}1 & -2 \\ 3 & -5 \\ -6 & 0\end{array}\right], B=\left[\begin{array}{cc}-1 & -2 \\ 4 & 2 \\ 1 & 5\end{array}\right]$ and $C=\left[\begin{array}{cc}2 & 4 \\ -1 & -4 \\ -3 & 6\end{array}\right]$, find the matrix $X$ such that $3 A-4 B+5 X=C$.
Answer$
\begin{aligned}
& 3 A-4 B+5 X=C \\
& \therefore 5 X=C-3 A+4 B \\
& =\left[\begin{array}{rr}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]-3\left[\begin{array}{rr}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]+4\left[\begin{array}{rr}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right] \\
& =\left[\begin{array}{rr}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]-\left[\begin{array}{rr}
3 & -6 \\
9 & -15 \\
-18 & 0
\end{array}\right]+\left[\begin{array}{rr}
-4 & -8 \\
16 & 8 \\
4 & 20
\end{array}\right] \\
& =\left[\begin{array}{rr}
2-3+(-4) & 4-(-6)-8 \\
-1-9+16 & -4-(-15)+8 \\
-3-(-18)+4 & 6-0+20
\end{array}\right] \\
& =\left[\begin{array}{rr}
-5 & 2 \\
6 & 19 \\
19 & 26
\end{array}\right] \\
& \therefore X=\frac{1}{5}\left[\begin{array}{rr}
-5 & 2 \\
6 & 19 \\
19 & 26
\end{array}\right]=\left[\begin{array}{rr}
-1 & \frac{2}{5} \\
6 & \frac{19}{5} \\
5 & \frac{26}{5} \\
\frac{19}{5} & 5
\end{array}\right]
\end{aligned}
$
View full question & answer→Question 253 Marks
Find $x$ and $y$, if
$
\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]
$
Answer$
\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{rrr}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]
$
$
\begin{aligned}
& \therefore\left[\begin{array}{lrr}
2 x+y-1 & -1+6 & 1+4 \\
3+3 & 4 y+0 & 4+3
\end{array}\right]=\left[\begin{array}{rrr}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right] \\
& \therefore\left[\begin{array}{lrr}
2 x+y-1 & 5 & 5 \\
6 & 4 y & 7
\end{array}\right]=\left[\begin{array}{rrr}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]
\end{aligned}
$
By equality of matrices, we get
$
2 x+y-1=3
$
and $4 y =18$
From (2), $y=\frac{9}{2}$
Substituting $y=\frac{9}{2}$ in (1), we get
$
\begin{aligned}
& 2 x+\frac{9}{2}-1=3 \\
& \therefore 2 x=3-\frac{7}{2}=\frac{-1}{2} \\
& \therefore x=\frac{-1}{4}
\end{aligned}
$
Hence, $x=\frac{-1}{4}$ and $y=\frac{9}{2}$.
View full question & answer→Question 263 Marks
Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular
1. $\left[\begin{array}{ccc}2 & 0 & 0 \\ 3 & -1 & 0 \\ -7 & 3 & 1\end{array}\right]$
2. $\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right]$
3. $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Answer(i) Since, all the elements above the diagonal are zero, it is a lower triangular matrix.
(ii) Since, all the non-diagonal elements are zero, it is a diagonal matrix.
(iii) Since, diagonal elements are 1 and non-diagonal elements are 0, it is an identity (or unit) matrix.
View full question & answer→