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Maths Solve the Following Question.(3 Marks)

Determinants and Matrices · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 50 questions.

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Solve the Following Question.(3 Marks)

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Question 13 Marks
Prove The Theorem : If $\mathrm{A}\left(x_1, y_1\right), \mathrm{B}\left(x_2, y_2\right)$ and $\mathrm{C}\left(x_3, y_3\right)$ are vertices of triangle $A B C$ then the area of triangle is
$
\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|
$
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Question 23 Marks
If $A=\left[\begin{array}{cc}2 & -4 \\ 3 & -2 \\ 0 & 1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 1 & 0\end{array}\right]$, show that $(A B)^{\top}=B^{\top} A^{\top}$.
Answer
$\begin{aligned} A B & =\left[\begin{array}{cc}2 & -4 \\ 3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 1 & 0\end{array}\right] \\ & =\left[\begin{array}{ccc}2+8 & -2-4 & 4+0 \\ 3+4 & -3-2 & 6-0 \\ 0-2 & 0+1 & 0+0\end{array}\right]\end{aligned}$

$=\left[\begin{array}{ccc}10 & -6 & 4 \\ 7 & -5 & 6 \\ -2 & 1 & 0\end{array}\right]$

$\therefore \quad(A B)^T=\left[\begin{array}{ccc}10 & 7 & -2 \\ -6 & -5 & 1 \\ 4 & 6 & 0\end{array}\right]$

$\ldots$...(i)

Now, $A^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 3 & 0 \\ -4 & -2 & 1\end{array}\right]$ and $B^{\mathrm{T}}=\left[\begin{array}{cc}1 & -2 \\ -1 & 1 \\ 2 & 0\end{array}\right]$

$\begin{aligned} \therefore \quad \mathbf{B}^{\top} \mathbf{A}^{\top} & =\left[\begin{array}{cc}1 & -2 \\ -1 & 1 \\ 2 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 3 & 0 \\ -4 & -2 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}2+8 & 3+4 & 0-2 \\ -2-4 & -3-2 & 0+1 \\ 4-0 & 6-0 & 0+0\end{array}\right]\end{aligned}$

$\therefore \quad \mathbf{B}^{\mathrm{T}} \mathbf{A}^{\mathrm{T}}=\left[\begin{array}{ccc}10 & 7 & -2 \\ -6 & -5 & 1 \\ 4 & 6 & 0\end{array}\right]$

...(ii)

From (i) and (ii), we get

$(A B)^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$

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Question 33 Marks
If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0\end{array}\right]$, show that $A B$ and $B A$ are both singular

matrices.

Answer
$\mathrm{AB}=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ccc}1-6-6 & -1+4+3 & 1-2+0 \\ 2-12-12 & -2+8+6 & 2-4+0 \\ 1-6-6 & -1+4+3 & 1-2+0\end{array}\right] \\ & =\left[\begin{array}{ccc}-11 & 6 & -1 \\ -22 & 12 & -2 \\ -11 & 6 & -1\end{array}\right]\end{aligned}$

$|A B|=\left|\begin{array}{ccc}-11 & 6 & -1 \\ -22 & 12 & -2 \\ -11 & 6 & -1\end{array}\right|$

$=0 \quad \ldots\left[\because R_1\right.$ an $R_3$ are identical $]$

$\mathrm{AB}$ is a singular matrix

$\begin{aligned} \mathbf{B A} & =\left[\begin{array}{ccc}1 & -1 & 1 \\ -3 & 2 & -1 \\ -2 & 1 & 0\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}1-2+1 & 2-4+2 & 3-6+3 \\ -3+4-1 & -6+8-2 & -9+12-3 \\ -2+2+0 & -4+4+0 & -6+6+0\end{array}\right] \\ & =\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\end{aligned}$

$|B A|=0$

BA is a singular matrix.

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Question 43 Marks
Show that the lines x – y = 6, 4x – 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also, find the point of concurrence.
Answer
Given equations of the lines are x – y = 6, i.e., x – y – 6 = 0 ……(i) 4x – 3y = 20, i.e., 4x – 3y – 20 = 0 …..(ii) 6x + 5y + 8 = 0 ……(iii) The given lines will be concurrent, if

$\begin{aligned} & \left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0 \\ & \left|\begin{array}{ccc}1 & -1 & -6 \\ 4 & -3 & -20 \\ 6 & 5 & 8\end{array}\right|\end{aligned}$

= 1(-24 + 100) – (-1) (32 + 120) – 6(20 + 18) = 1(76) + 1(152) – 6(38) = 76 + 152 – 228 = 0 ∴ The given lines are concurrent. To find the point of concurrence, solve any two equations. Multiplying (i) by 5, we get 5x – 5y – 30 = 0 …….(iv) Adding (iii) and (iv), we get 11x – 22 = 0 ∴ x = 2 Substituting x = 2 in (i), we get 2 – y – 6 = 0 ∴ y = -4 ∴ The point of concurrence is (2, -4).

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Question 53 Marks
Find the area of quadrilateral whose vertices are A(0, -4), B(4, 0), C(-4,0), D (0, 4).
Answer
A(0, -4), B(4, 0), C(-4, 0), D(0, 4)

Image

$\mathrm{A}(\square \mathrm{ABDC})=\mathrm{A}(\triangle \mathrm{ABC})+\mathrm{A}(\triangle \mathrm{BDC})$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\mathrm{A}(\Delta \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{ccc}0 & -4 & 1 \\ 4 & 0 & 1 \\ -4 & 0 & 1\end{array}\right|$

$=\frac{1}{2}[0-(-4)(4+4)+1(0-0)]$

$=\frac{1}{2}(32)=16$ sq. units

$\mathrm{A}(\Delta \mathrm{BDC})=\frac{1}{2}\left|\begin{array}{ccc}4 & 0 & 1 \\ 0 & 4 & 1 \\ -4 & 0\end{array}\right|$

$=\frac{1}{2}[4(4-0)-0+1(0+16)]$

$=\frac{1}{2}[4(4)+1(16)]$

$=\frac{1}{2}(16+16)$

$=\frac{1}{2}(32)$

$=16 \mathrm{sq}$. units

∴ A(ABDC) = A(ΔABC) + A(ΔBDC) = 16 + 16 = 32 sq.units

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Question 63 Marks
If $\left|\begin{array}{ccc}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$ then show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$
Answer
$\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$

Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get

$\left|\begin{array}{ccc}a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1\end{array}\right|=0$

$\begin{aligned} \therefore \quad a[(b-1)(c-1)-0]- & 1[(1-a)(c-1)-0] \\ & +1[0-(b-1)(1-a)]=0\end{aligned}$

$\begin{aligned} \therefore \quad a(1-b)(1-c)+(1-a)(1-c) \\ +(1-b)(1-a)=0\end{aligned}$

Dividing throughout by $(1-a)(1-b)(1-c)$,

we get

$\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$

Adding 1 on both the sides, we get

$1+\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$

$\therefore \quad \frac{1-a+a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$

$\therefore \quad \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$

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Question 73 Marks
Without expanding the determinants, show that

$\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|=0$

Answer
Let $D=\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|$

Taking (-1) common from $R_1, R_2, R_3$, we get

$D=(-1)^3\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|$

Interchanging rows and columns, we get

$D=-1\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|$

$\begin{array}{ll}\therefore & D=-1(D) \\ \therefore & 2 D=0 \\ \therefore & D=0\end{array}$

$\therefore \quad\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|=0$

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Question 83 Marks
Without expanding the determinants, show that

$\left|\begin{array}{lll}l & m & n \\ e & d & f \\ u & v & w\end{array}\right|=\left|\begin{array}{lll}n & f & w \\ l & e & u \\ m & d & v\end{array}\right|$

Answer
L.H.S. $=\left|\begin{array}{lll}l & \mathrm{~m} & \mathrm{n} \\ \mathrm{e} & \mathrm{d} & \mathrm{f} \\ \mathrm{u} & \mathbf{v} & \mathrm{w}\end{array}\right|$

Interchanging rows and columns, we get

L.H.S. $=\left|\begin{array}{lll}l & \mathrm{e} & \mathrm{u} \\ \mathrm{m} & \mathrm{d} & \mathrm{v} \\ \mathrm{n} & \mathrm{f} & \mathrm{w}\end{array}\right|$

Applying $\mathrm{R}_2 \leftrightarrow \mathrm{R}_3$, we get

L.H.S. $=-\left|\begin{array}{lll}l & e & u \\ n & f & w \\ m & d & v\end{array}\right|$

Applying $R_1 \leftrightarrow R_2$, we get

$\begin{aligned} \text { L.H.S. } & =\left|\begin{array}{ccc}\mathrm{n} & \mathrm{f} & \mathrm{w} \\ l & \mathrm{e} & \mathrm{u} \\ \mathrm{m} & \mathrm{d} & \mathrm{v}\end{array}\right| \\ & =\text { R.H.S. }\end{aligned}$

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Question 93 Marks
Without expanding the determinants, show that

$\left|\begin{array}{ccc}x a & y b & z c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|$$=\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ b c & c a & a b\end{array}\right|$

Answer
L.H.S. $=\left|\begin{array}{ccc}x a & y b & z c \\ \mathrm{a}^2 & \mathrm{~b}^2 & \mathrm{c}^2 \\ 1 & 1 & 1\end{array}\right|$

Taking a, b, c common from $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3$

respectively, we get

L.H.S. $=a b c\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{array}\right|$

$=\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ \frac{a b c}{a} & \frac{a b c}{b} & \frac{a b c}{c}\end{array}\right|$

$=\left|\begin{array}{ccc}x & y & z \\ a & b & \mathrm{c} \\ \mathrm{bc} & c a & a b\end{array}\right|=$ R.H.S.

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Question 103 Marks
Without expanding the determinants, show that

$\left|\begin{array}{ccc}b+c & b c & b^2 c^2 \\ c+a & c a & c^2 a^2 \\ a+b & a b & a^2 b^2\end{array}\right|=0$

Answer
L.H.S. $=\left|\begin{array}{ccc}b+c & b c & b^2 c^2 \\ c+a & c a & c^2 a^2 \\ a+b & a b & a^2 b^2\end{array}\right|$

Taking bc, ca, ab common from $R_1, R_2, R_3$

respectively, we get

L.H.S. $=(b c)(c a)(a b)\left|\begin{array}{ccc}\frac{b+c}{b c} & 1 & b c \\ \frac{c+a}{c a} & 1 & c a \\ \frac{a+b}{a b} & 1 & a b\end{array}\right|$

Taking abc common from $\mathrm{C}_3$, we get

L.H.S. $=\left(\mathrm{a}^2 \mathrm{~b}^2 \mathrm{c}^2\right)(a b c)\left|\begin{array}{lll}\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{~b}} & 1 & \frac{1}{a} \\ \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}} & 1 & \frac{1}{b} \\ \frac{1}{b}+\frac{1}{\mathrm{a}} & 1 & \frac{1}{c}\end{array}\right|$

Applying $C_1 \rightarrow C_1+C_3$, we get

L.H.S. $=a^3 b^3 c^3\left|\begin{array}{lll}\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{a} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{b} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{c}\end{array}\right|$

Taking $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ common from $C_1$, we get

$=a^3 b^3 c^3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(0)$

$\ldots\left[\because C_1\right.$ and $C_2$ are identical $]$

$=0=$ R.H.S.

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Question 113 Marks
A school purchased 8 dozen Mathematics books, 7 dozen Physics books and 10 dozen
Chemistry books, the prices are Rs.50,Rs.40 and Rs.60 per book respectively. Find the total amount that the book seller will receive from school authority using matrix multiplication.
Answer
Let A be the column matrix of books of different subjects and let B be the row matrix of prices of one book of each subject.
$
\mathrm{A}=\left[\begin{array}{c}
8 \times 12 \\
7 \times 12 \\
10 \times 12
\end{array}\right]=\left[\begin{array}{c}
96 \\
84 \\
120
\end{array}\right] \quad \mathrm{B}=\left[\begin{array}{lll}
50 & 40 & 60
\end{array}\right]
$
$\therefore$ The total amount received by the bookseller is obtained by matrix BA.
$
\begin{aligned}
& \therefore \mathrm{BA}=\left[\begin{array}{lll}
50 & 40 & 60
\end{array}\right]\left[\begin{array}{c}
96 \\
84 \\
120
\end{array}\right] \\
& =[50 \times 96+40 \times 84+60 \times 120] \\
& =[4800+3360+7200]=[15360]
\end{aligned}
$
Thus the amount received by the bookseller from the school is Rs. 15360 .
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Question 123 Marks
If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$, Find $\mathrm{k}$, so that $A^2-\mathrm{kA}+2 \mathrm{I}$ $=\mathrm{O}$, where $\mathrm{I}$ is an identity matrix and $\mathrm{O}$ is null matrix of order 2 .
Answer
Given $A^2-\mathrm{kA}+2 \mathrm{I}=\mathrm{O}$
$
\begin{aligned}
& \therefore \text { Here, } A^2=\mathrm{AA}=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{rr}
3 & -2 \\
4 & -2
\end{array}\right] \\
& =\left[\begin{array}{rr}
9-8 & -6+4 \\
12-8 & -8+4
\end{array}\right]=\left[\begin{array}{rr}
1 & -2 \\
4 & -4
\end{array}\right] \\
& \therefore A^2-\mathrm{kA}+2 \mathrm{I}=\mathrm{O} \\
& \therefore\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]-\mathrm{k}\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]+2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{O} \\
& \therefore\left[\begin{array}{rr}
1 & -2 \\
4 & -4
\end{array}\right]-\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]+\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
1-3 k+2 & -2+2 k \\
4-4 k & -4+2 k+2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
$\therefore$ Using definition of equality of matrices, we have
$
\left.\begin{array}{lll}
1-3 \mathrm{k}+2=0 & \therefore & 3 \mathrm{k}=3 \\
-2+k=0 & \therefore & 2 \mathrm{k}=2 \\
4-4 k=0 & \therefore & 4 \mathrm{k}=4 \\
-4+2 \mathrm{k}+2=0 & \therefore & 2 \mathrm{k}=2
\end{array}\right\} \quad \mathrm{k}=1
$
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Question 133 Marks
If $\mathrm{A}=\left[\begin{array}{lll}1 & 3 & 3 \\ 3 & 1 & 3 \\ 3 & 3 & 1\end{array}\right]$ prove that $A^2-5 \mathrm{~A}$ is a scalar matrix. $\left[\begin{array}{lll}3 & 3 & 1\end{array}\right]$
Answer
Let $A^2=\mathrm{A}$.A
$
\begin{aligned}
& =\left[\begin{array}{lll}
1 & 3 & 3 \\
3 & 1 & 3 \\
3 & 3 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 3 & 3 \\
3 & 1 & 3 \\
3 & 3 & 1
\end{array}\right] \\
& =\left[\begin{array}{lll}
1+9+9 & 3+3+9 & 3+9+3 \\
3+3+9 & 9+1+9 & 9+3+3 \\
3+9+3 & 9+3+3 & 9+9+1
\end{array}\right] \\
& =\left[\begin{array}{lll}
19 & 15 & 15 \\
15 & 19 & 15 \\
15 & 15 & 19
\end{array}\right] \\
\therefore A^2-5 \mathrm{~A} & =\left[\begin{array}{lll}
19 & 15 & 15 \\
15 & 19 & 15 \\
15 & 15 & 19
\end{array}\right]-5\left[\begin{array}{lll}
1 & 3 & 3 \\
3 & 1 & 3 \\
3 & 3 & 1
\end{array}\right]
\end{aligned}
$

$
\begin{aligned}
& =\left[\begin{array}{lll}
19 & 15 & 15 \\
15 & 19 & 15 \\
15 & 15 & 19
\end{array}\right]-\left[\begin{array}{ccc}
5 & 15 & 15 \\
15 & 5 & 15 \\
15 & 15 & 5
\end{array}\right] \\
\therefore A^2-5 A & =\left[\begin{array}{ccc}
14 & 0 & 0 \\
0 & 14 & 0 \\
0 & 0 & 14
\end{array}\right]=14 \mathrm{I}
\end{aligned}
$
$\therefore A^2-5 \mathrm{~A}$ is a scalar matrix.

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Question 143 Marks
If $X+Y=\left[\begin{array}{cc}2 & -1 \\ 1 & 3 \\ -3 & -2\end{array}\right]$ and
$\mathrm{X}-2 \mathrm{Y}$
$=\left[\begin{array}{cc}-2 & 1 \\ 3 & -1 \\ 4 & -2\end{array}\right]$ then find $\mathrm{X}, \mathrm{Y}$.
Answer
Let $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 3 \\ -3 & -2\end{array}\right]$ and $B=\left[\begin{array}{cc}-2 & 1 \\ 3 & -1 \\ 4 & -2\end{array}\right]$
Let, $\mathrm{X}+\mathrm{Y}=\mathrm{A} \ldots$ (1), $\mathrm{X}-2 \mathrm{Y}=\mathrm{B}$
Solving (1) and (2) for $\mathrm{X}$ and $\mathrm{Y}$

$\begin{aligned} & \text { By }(1)-(2), 3 Y=A-B, \therefore Y=\frac{1}{3}(A-B) \\ & \therefore Y=\frac{1}{3}\left\{\left[\begin{array}{cc}2 & -1 \\ 1 & 3 \\ -3 & -2\end{array}\right]-\left[\begin{array}{cc}-2 & 1 \\ 3 & -1 \\ 4 & -2\end{array}\right]\right\}=\frac{1}{3}\left[\begin{array}{cc}4 & -2 \\ -2 & 4 \\ -7 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}\frac{4}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{4}{3} \\ -\frac{7}{3} & 0\end{array}\right] \\ & \text { From (1) } \mathrm{X}+\mathrm{Y}=\mathrm{A}, \quad \therefore \mathrm{X}=\mathrm{A}-\mathrm{Y} \text {, } \\ & \therefore X=\left[\begin{array}{cc}2 & -1 \\ 1 & 3 \\ -3 & -2\end{array}\right]-\left[\begin{array}{cc}\frac{4}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{4}{3} \\ -\frac{7}{3} & 0\end{array}\right] \\ & X=\left[\begin{array}{cc}2-\frac{4}{3} & -1+\frac{2}{3} \\ 1+\frac{2}{3} & 3-\frac{4}{3} \\ -3+\frac{7}{3} & -2+0\end{array}\right]=\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{5}{3} & \frac{5}{3} \\ -\frac{2}{3} & -2\end{array}\right] \\ & \end{aligned}$

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Question 153 Marks
If $\left|\begin{array}{lll}x & y & z \\ -x & y & z \\ x & -y & z\end{array}\right|=\begin{aligned} & k x y z \text { then find the value } \\ & \text { of } \mathrm{k}\end{aligned}$
Answer
$
\begin{aligned}
& \text { L.H.S. }=\left|\begin{array}{lll}
x & y & z \\
-x & y & z \\
x & -y & z
\end{array}\right| \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_1 \\
& =\left|\begin{array}{rrr}
x & y & z \\
0 & 2 y & 2 z \\
x & -y & z
\end{array}\right| \\
& \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_1 \\
& =\left|\begin{array}{lll}
x & y & z \\
0 & 2 y & 2 z \\
2 x & 0 & 2 z
\end{array}\right| \\
& =2 \times 2\left|\begin{array}{lll}
x & y & z \\
0 & y & z \\
x & 0 & z
\end{array}\right| \text { taking (2 common } \\
& \text { from } R_2 \text { and } R_3 \text { ) } \\
& =4[x(y z)-y(0-x z)+z(0-x y)] \\
& =4[x y z+x y z-x y z] \\
& =4 x y z \\
&
\end{aligned}
$
From given condition
L.H.S. $=$ R.H.S.
$
\begin{aligned}
4 x y z & =k x y z \\
\therefore k & =4
\end{aligned}
$
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Question 163 Marks
$\left|\begin{array}{lll}312 & 313 & 314 \\ 315 & 316 & 317 \\ 318 & 319 & 320\end{array}\right|=0$
Answer
$
\begin{aligned}
\text { L.H.S. }= & \left|\begin{array}{lll}
312 & 313 & 314 \\
315 & 316 & 317 \\
318 & 319 & 320
\end{array}\right| \\
& C_2 \rightarrow C_2-C_1 \\
= & \left|\begin{array}{lll}
312 & 1 & 314 \\
315 & 1 & 317 \\
318 & 1 & 320
\end{array}\right| \\
& C_3 \rightarrow C_3-C_1 \\
= & \left|\begin{array}{lll}
312 & 1 & 2 \\
315 & 1 & 2 \\
318 & 1 & 2
\end{array}\right|
\end{aligned}
$
take 2 common from $\mathrm{C}_3$
$
\begin{aligned}
& =2\left|\begin{array}{lll}
312 & 1 & 1 \\
315 & 1 & 1 \\
318 & 1 & 1
\end{array}\right| \\
& =2(0) \quad\left(\mathrm{C}_2 \text { and } \mathrm{C}_3 \text { are identical }\right)
\end{aligned}
$
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Question 173 Marks
Evaluate : $\left|\begin{array}{ccc}2-i & 3 & -1 \\ 3 & 2-i & 0 \\ 2 & -1 & 2-i\end{array}\right|$ where $i=\sqrt{-1}$
Answer
$\begin{aligned}& \left|\begin{array}{ccc}2-i & 3 & -1 \\ 3 & 2-i & 0 \\ 2 & -1 & 2-i\end{array}\right| \\ & =(2-i)\left[(2-i)^2-0\right]-3[3(2-i)-0] \\ & -1[-3-2(2-\mathrm{i})] \\ & =(2-i)^3-9(2-i)+3+2(2-i) \\ & =8-12 i+6 i^2-i^3-18+9 i+3+4-2 i \\ & =8-12 i+6(-1)+i-18+9 i+3+4-2 i \\ & \text { (since } i^2=-1 \text { ) } \\ & =8-6-18+7-12 i+9 i-2 i+i \\ & =-9-4 i \\ & \end{aligned}$
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Question 183 Marks
Evaluate : $\left|\begin{array}{lll}\sec \theta & \tan \theta & 0 \\ \tan \theta & \sec \theta & 0 \\ 0 & 0 & 1\end{array}\right|$
Answer
$\begin{aligned} & \left|\begin{array}{lll}\sec \theta & \tan \theta & 0 \\ \tan \theta & \sec \theta & 0 \\ 0 & 0 & 1\end{array}\right| \\ & =\sec \theta\left|\begin{array}{cc}\sec \theta & 0 \\ 0 & 1\end{array}\right|-\tan \theta\left|\begin{array}{cc}\tan \theta & 0 \\ 0 & 1\end{array}\right| \\ & +0\left|\begin{array}{cc}\tan \theta & \sec \theta \\ 0 & 0\end{array}\right| \\ & =\sec \theta(\sec \theta-0)-\tan \theta(\tan \theta-0)+0 \\ & =\sec ^2 \theta-\tan ^2 \theta \\ & =1 \end{aligned}$
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Question 193 Marks
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.

$\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$

Answer
Let $A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$

$\therefore \quad A^T=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

$\therefore \quad A+A^{\mathrm{T}}=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

$=\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]$

Also, $A-A^T=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

$=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^T\right)=\left[\begin{array}{ccc}3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2\end{array}\right]$

and $Q=\frac{1}{2}\left(A-A^T\right)=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ \frac{-5}{2} & 0 & 3 \\ \frac{-3}{2} & -3 & 0\end{array}\right]$

$\therefore P$ is symmetric matrix $\ldots\left[\because a_{-1}=a_{j-1}\right]$

and $Q$ is a skew symmetric matrix $\left[\because-a_{i j}=-a_{j i j}\right]$

∴ A = P + Q

$\therefore A=\left[\begin{array}{cc}4 & \frac{1}{2} \\ \frac{1}{2} & -5\end{array}\right]+\left[\begin{array}{cc}0 & \frac{-5}{2} \\ \frac{5}{2} & 0\end{array}\right]$

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Question 203 Marks
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.

$\left[\begin{array}{rr}4 & -2 \\ 3 & -5\end{array}\right]$

Answer
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as

$A=\frac{1}{2}\left(A+A^{\top}\right)+\frac{1}{2}\left(A-A^{\top}\right)$

i. Let $A=\left[\begin{array}{ll}4 & -2 \\ 3 & -5\end{array}\right]$

$\therefore \quad A^{\mathrm{T}}=\left[\begin{array}{cc}4 & 3 \\ -2 & -5\end{array}\right]$

$\therefore \quad \mathrm{A}+\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ll}4 & -2 \\ 3 & -5\end{array}\right]+\left[\begin{array}{cc}4 & 3 \\ -2 & -5\end{array}\right]$

$=\left[\begin{array}{cc}8 & 1 \\ 1 & -10\end{array}\right]$

Also, $A-A^T=\left[\begin{array}{ll}4 & -2 \\ 3 & -5\end{array}\right]-\left[\begin{array}{cc}4 & 3 \\ -2 & -5\end{array}\right]$

$=\left[\begin{array}{cc}0 & -5 \\ 5 & 0\end{array}\right]$

Let $\mathrm{P}=\frac{1}{2}\left(A+A^T\right)=\left[\begin{array}{cc}4 & \frac{1}{2} \\ \frac{1}{2} & -5\end{array}\right]$

and $Q=\frac{1}{2}\left(A-A^T\right)=\left[\begin{array}{cc}0 & -\frac{5}{2} \\ \frac{5}{2} & 0\end{array}\right]$

$P$ is symmetric matrix $\ldots\left[\because a_{\mathrm{ij}}=a_{\mathrm{j}}\right]$

and $Q$ is a skew symmetric matrix $\left[\because-a_{\mathrm{ij}}=-\mathrm{a}_{\mathrm{j}}\right]$

A = P + Q

$A=\left[\begin{array}{cc}4 & \frac{1}{2} \\ \frac{1}{2} & -5\end{array}\right]+\left[\begin{array}{cc}0 & \frac{-5}{2} \\ \frac{5}{2} & 0\end{array}\right]$

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Question 213 Marks
If $A=\left[\begin{array}{ccc}7 & 3 & 0 \\ 0 & 4 & -2\end{array}\right], B=\left[\begin{array}{ccc}0 & -2 & 3 \\ 2 & 1 & -4\end{array}\right]$, then find

i. $A^{\top}+4 B^{\top}$

ii. $5 A^{\top}-5 B^{\top}$

Answer
$A=\left[\begin{array}{ccc}7 & 3 & 0 \\ 0 & 4 & -2\end{array}\right]$ and $B=\left[\begin{array}{ccc}0 & -2 & 3 \\ 2 & 1 & -4\end{array}\right]$

$\therefore \quad A^T=\left[\begin{array}{cc}7 & 0 \\ 3 & 4 \\ 0 & -2\end{array}\right]$ and $B^T=\left[\begin{array}{cc}0 & 2 \\ -2 & 1 \\ 3 & -4\end{array}\right]$

$A^T+4 B^T=\left[\begin{array}{cc}7 & 0 \\ 3 & 4 \\ 0 & -2\end{array}\right]+4\left[\begin{array}{cc}0 & 2 \\ -2 & 1 \\ 3 & -4\end{array}\right]$

$=\left[\begin{array}{cc}7 & 0 \\ 3 & 4 \\ 0 & -2\end{array}\right]+\left[\begin{array}{cc}0 & 8 \\ -8 & 4 \\ 12 & -16\end{array}\right]$

$=\left[\begin{array}{cc}7+0 & 0+8 \\ 3-8 & 4+4 \\ 0+12 & -2-16\end{array}\right]=\left[\begin{array}{cc}7 & 8 \\ -5 & 8 \\ 12 & -18\end{array}\right]$

$5 A^{\top}-5 B^{\top}=5\left(A^{\top}-B^{\top}\right)$

$=5\left(\left[\begin{array}{cc}7 & 0 \\ 3 & 4 \\ 0 & -2\end{array}\right]-\left[\begin{array}{cc}0 & 2 \\ -2 & 1 \\ 3 & -4\end{array}\right]\right)$

$=5\left[\begin{array}{cc}7-0 & 0-2 \\ 3+2 & 4-1 \\ 0-3 & -2+4\end{array}\right]=5\left[\begin{array}{cc}7 & -2 \\ 5 & 3 \\ -3 & 2\end{array}\right]=\left[\begin{array}{cc}35 & -10 \\ 25 & 15 \\ -15 & 10\end{array}\right]$

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Question 223 Marks
If $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 1 \\ 4 & -1 \\ -3 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$ then show that

$(A-C)^{\top}=A^{\top}-C^{\top}$

Answer
$A-C=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]-\left[\begin{array}{cc}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$

$=\left[\begin{array}{cc}2-1 & -3-2 \\ 5+1 & -4-4 \\ -6+2 & 1-3\end{array}\right]$

$=\left[\begin{array}{cc}1 & -5 \\ 6 & -8 \\ -4 & -2\end{array}\right]$

$\therefore \quad(A-C)^T=\left[\begin{array}{ccc}1 & 6 & -4 \\ -5 & -8 & -2\end{array}\right]$

$\ldots(\mathrm{i})$

Now, $A^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 5 & -6 \\ -3 & -4 & 1\end{array}\right]$ and

$C^T=\left[\begin{array}{ccc}1 & -1 & -2 \\ 2 & 4 & 3\end{array}\right]$

$\therefore \quad A^T-C^T=\left[\begin{array}{ccc}2 & 5 & -6 \\ -3 & -4 & 1\end{array}\right]-\left[\begin{array}{ccc}1 & -1 & -2 \\ 2 & 4 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}2-1 & 5+1 & -6+2 \\ -3-2 & -4-4 & 1-3\end{array}\right]$

$=\left[\begin{array}{ccc}1 & 6 & -4 \\ -5 & -8 & -2\end{array}\right]$

...(ii)

From (i) and (ii), we get

$(A-C)^{\top}=A^{\top}-C^{\top}$

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Question 233 Marks
If $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 1 \\ 4 & -1 \\ -3 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$ then show that

i. $(A+B)^{\top}=A T+B^{\top}$

Answer
$A+B=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]+\left[\begin{array}{cc}2 & 1 \\ 4 & -1 \\ -3 & 3\end{array}\right]$

$=\left[\begin{array}{cc}2+2 & -3+1 \\ 5+4 & -4-1 \\ -6-3 & 1+3\end{array}\right]=\left[\begin{array}{cc}4 & -2 \\ 9 & -5 \\ -9 & 4\end{array}\right]$

$\therefore \quad(A+B)^T=\left[\begin{array}{ccc}4 & 9 & -9 \\ -2 & -5 & 4\end{array}\right]$

$\ldots$ (i)

Now, $\mathbf{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 5 & -6 \\ -3 & -4 & 1\end{array}\right]$ and $\mathbf{B}^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 4 & -3 \\ 1 & -1 & 3\end{array}\right]$

$\therefore \quad \mathrm{A}^{\mathrm{T}}+\mathbf{B}^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 5 & -6 \\ -3 & -4 & 1\end{array}\right]+\left[\begin{array}{ccc}2 & 4 & -3 \\ 1 & -1 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}2+2 & 5+4 & -6-3 \\ -3+1 & -4-1 & 1+3\end{array}\right]$

$=\left[\begin{array}{ccc}4 & 9 & -9 \\ -2 & -5 & 4\end{array}\right]$

...(ii)

From (i) and (ii), we get

$(A+B)^{\top}=A T+B^{\top}$

[Note: The question has been modified.]

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Question 243 Marks
If $\left[a_{i-1}\right]_{3 \times 3}$ where $a_{i j}=2(i-j)$, find $A$ and

$A^{\top}$. State whether $A$ and $A^{\top}$ are symmetric or skew-symmetric matrices?

Answer
$\mathrm{A}=\left[a_{\mathrm{ij}}\right]_{3 \times 3}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$

Given $\mathrm{a}_{\mathrm{ij}}=2(\mathrm{i}-\mathrm{j})$

$\begin{aligned} & \therefore a_{11}=2(1-1)=0, \\ & a_{12}=2(1-2)=-2, \\ & a_{13}=2(1-3)=-4 \\ & a_{21}=2(2-1)=2,\end{aligned}$

$\begin{aligned} & a_{22}=2(2-2)=0 \\ & a_{23}=2(2-3)=-2 \\ & a_{31}=2(3-1)=4 \\ & a_{32}=2(3-2)=2 \\ & a_{33}=2(3-3)=0\end{aligned}$

$\therefore \quad \mathrm{A}=\left[\begin{array}{ccc}0 & -2 & -4 \\ 2 & 0 & -2 \\ 4 & 2 & 0\end{array}\right]$

$\therefore \quad A^T=\left[\begin{array}{ccc}0 & 2 & 4 \\ -2 & 0 & 2 \\ -4 & -2 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & -2 & -4 \\ 2 & 0 & -2 \\ 4 & 2 & 0\end{array}\right]=-A$

$\therefore A^{\top}=-A$ and $A=-A^{\top}$

$\therefore A$ and $\mathrm{A}^{\top}$ both are skew-symmetric matrices.

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Question 253 Marks
Jay and Ram are two friends in a class. Jay wanted to buy $4$ pens and $8$ notebooks, Ram wanted to buy $5$ pens and $12$ notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was $6$ and $₹ 10.$ Using matrix multiplication, find the amount required from each one of them.
Answer
Let $A$ be the matrix of pens and notebooks and $B$ be the matrix of prices of one pen and one notebook.
Pens Notebooks
$\therefore \mathrm{A}=\left[\begin{array}{lc}4 & 8 \\ 5 & 12\end{array}\right]$ Jay
and $B=\left[\begin{array}{c}6 \\ 10\end{array}\right] \begin{aligned} & \text { Pen } \\ & \text { Notebook }\end{aligned}$
The total amount required for each one of them is obtained by matrix $AB.$
$\therefore \mathrm{AB}=\left[\begin{array}{cc}4 & 8 \\ 5 & 12\end{array}\right]\left[\begin{array}{c}6 \\ 10\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{l}24+80 \\ 30+120\end{array}\right]  \end{aligned} $
$=\left[\begin{array}{l}104 \\ 150\end{array}\right]$
$\therefore$ Jay needs $₹ 104$ and Ram needs $₹ 150.$
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Question 263 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right], B=\left[\begin{array}{cc}0 & 4 \\ 2 & -1\end{array}\right]$

show that AB ≠ BA, but |AB| = |A| . |B|.

Answer
$\begin{aligned} A B & =\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right]\left[\begin{array}{cc}0 & 4 \\ 2 & -1\end{array}\right] \\ & =\left[\begin{array}{cc}0+4 & 4-2 \\ 0+10 & 12-5\end{array}\right]\end{aligned}$

$=\left[\begin{array}{cc}4 & 2 \\ 10 & 7\end{array}\right]$

$\begin{aligned} \mathbf{B A} & =\left[\begin{array}{cc}0 & 4 \\ 2 & -1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 5\end{array}\right] \\ & =\left[\begin{array}{cc}0+12 & 0+20 \\ 2-3 & 4-5\end{array}\right] \\ & =\left[\begin{array}{cc}12 & 20 \\ -1 & -1\end{array}\right] \neq \mathrm{AB}\end{aligned}$

Now $_r|A B|=\left|\begin{array}{cc}4 & 2 \\ 10 & 7\end{array}\right|=28-20=8$

$\begin{aligned} & |A|=\left|\begin{array}{cc}1 & 2 \\ 3 & 5\end{array}\right|=5-6=-1 \\ & |B|=\left|\begin{array}{cc}0 & 4 \\ 2 & -1\end{array}\right|=0-8=-8\end{aligned}$

∴ |A| . |B| = (-1).(-8) = 8 = |AB| ∴ AB ≠ BA, but |AB| = |A|.|B|

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Question 273 Marks
$\left\{3\left[\begin{array}{ll}2 & 0 \\ 0 & 2 \\ 2 & 2\end{array}\right]-4\left[\begin{array}{cc}1 & 1 \\ -1 & 2 \\ 3 & 1\end{array}\right]\right\}\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
Answer
$\therefore\left\{\left[\begin{array}{ll}6 & 0 \\ 0 & 6 \\ 6 & 6\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ -4 & 8 \\ 12 & 4\end{array}\right]\right\}\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore\left[\begin{array}{ll}6-4 & 0-4 \\ 0+4 & 6-8 \\ 6-12 & 6-4\end{array}\right]\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore\left[\begin{array}{rr}2 & -4 \\ 4 & -2 \\ -6 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore\left[\begin{array}{c}2-8 \\ 4-4 \\ -6+4\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore \left[\begin{array}{c}-6 \\ 0 \\ -2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore$ By equality of matrices,
we get $x-3=-6, y-1=0,2 z=-2 $
$\therefore x=-3, y=1, z=-1$
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Question 283 Marks
Find $x$ and $y$, if $\left\{4\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & 1 & 1\end{array}\right]\right\}\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
Answer
$\begin{aligned} & \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]\end{aligned}  $
$ \therefore\left\{\left[\begin{array}{ccc}8 & -4 & 12 \\ 4 & 0 & 8\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]$
$\begin{aligned} & \therefore \left[\begin{array}{ccc}8-3 & -4+3 & 12-4 \\ 4-2 & 0-1 & 8-1\end{array}\right]\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right] \end{aligned} $
$ \therefore \left[\begin{array}{lll}5 & -1 & 8 \\ 2 & -1 & 7\end{array}\right]\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\begin{aligned} & \therefore \left[\begin{array}{c}10+1+8 \\ 4+1+7\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right] \end{aligned} $
$ \therefore \left[\begin{array}{l}19 \\ 12\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\therefore$ By equality of matrices, we get $\mathrm{x}=19$ and $\mathrm{y}=12$
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Question 293 Marks
Find matrix $X$ such that $AX = B,$ where $A=\left[\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right]$ and $B=\left[\begin{array}{c}-3 \\ -1\end{array}\right]$
Answer
Let $\mathrm{X}=\left[\begin{array}{l}\boldsymbol{a} \\ b\end{array}\right]$ But $AX = B$
$\begin{aligned} & \therefore\left[\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{l}-3 \\ -1\end{array}\right] \end{aligned} $
$ \therefore\left[\begin{array}{c}a-2 b \\ -2 a+b\end{array}\right]=\left[\begin{array}{l}-3 \\ -1\end{array}\right]$
By equality of matrices, we get $a – 2b = -3 …(i) -2a + b = -l …(ii)$
By $(i) x ^2 + (ii),$ we get $-3b =-7$
$\therefore b=\frac{7}{3}$
Substituting $b=\frac{7}{3}$ in $(i),$ we get
$a-2\left(\frac{7}{3}\right)=-3$
$\therefore a=-3+\frac{14}{3}=\frac{5}{3}$
$\therefore X=\left[\begin{array}{l}\frac{5}{3} \\ \frac{7}{3}\end{array}\right]$
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Question 303 Marks
If $A=\left[\begin{array}{cc}1 & 2 \\ -1 & -2\end{array}\right], B=\left[\begin{array}{cc}2 & a \\ -1 & b\end{array}\right]$ and $(A+B)^2=A^2+B^2$, find the values of $a$ and $b$.
Answer
Given, $(A+B)^2=A^2+B^2$

$\therefore \mathrm{A}^2+\mathrm{AB}+\mathrm{BA}+\mathrm{B}^2=\mathrm{A}^2+\mathrm{B}^2$

∴ AB + BA = 0 ∴ AB = -BA

$\therefore \quad\left[\begin{array}{cc}1 & 2 \\ -1 & -2\end{array}\right]\left[\begin{array}{cc}2 & a \\ -1 & b\end{array}\right]=-\left[\begin{array}{cc}2 & a \\ -1 & b\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -1 & -2\end{array}\right]$

$\therefore \quad\left[\begin{array}{cc}2-2 & a+2 b \\ -2+2 & -a-2 b\end{array}\right]=-\left[\begin{array}{cc}2-a & 4-2 a \\ -1-b & -2-2 b\end{array}\right]$

∴ by equality of matrices, we get – 2 + a = 0 and 1 + b = 0 a = 2 and b = -1 [Note: The question has been modified.]

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Question 313 Marks
If $A=\left[\begin{array}{cc}3 & 4 \\ -4 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ -1 & 2\end{array}\right]$, show that $(A+B)(A-B)=A^2-B^2$.
Answer
We have to prove that $(A+B)(A-B)=A^2-B^2$.

i.e, to prove $A^2-A B+B A-B^2=A^2-B^2$

i.e, to prove $-\mathrm{AB}+\mathrm{BA}=0$,

i.e., to prove $A B-B A$.

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

$=\left[\begin{array}{cc}6-4 & 3+8 \\ -8-3 & -4+6\end{array}\right]$

$=\left[\begin{array}{cc}2 & 11 \\ -11 & 2\end{array}\right]$

$\ldots$..(i)

$\mathbf{B A}=\left[\begin{array}{cc}2 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 4 \\ -4 & 3\end{array}\right]$

$=\left[\begin{array}{cc}6-4 & 8+3 \\ -3-8 & -4+6\end{array}\right]$

$=\left[\begin{array}{cc}2 & 11 \\ -11 & 2\end{array}\right]$

$\ldots$..(ii)

From (i) and (ii), we get AB = BA

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Question 323 Marks
If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$, find $\mathrm{k}$ so that $\mathrm{A}^2-8 \mathrm{~A}-\mathrm{kl}=\mathrm{O}$, where $\mathrm{I}$ is a unit matrix and $\mathrm{O}$ is a null

matrix of order 2.

Answer
$\begin{aligned} & \mathrm{A}^2-8 \mathrm{~A}-\mathrm{kl}=0 \\ & \therefore \mathrm{A} \cdot \mathrm{A}-8 \mathrm{~A}-\mathrm{kl}=0\end{aligned}$

$\begin{aligned} \therefore\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right]-8\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right] & -\mathrm{k}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}$

$\begin{array}{ll}\therefore & {\left[\begin{array}{rr}1+0 & 0+0 \\ -1-7 & 0+49\end{array}\right]-\left[\begin{array}{rr}8 & 0 \\ -8 & 56\end{array}\right]-\left[\begin{array}{ll}\mathrm{k} & 0 \\ 0 & \mathrm{k}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]} \\ \therefore & {\left[\begin{array}{rc}1 & 0 \\ -8 & 49\end{array}\right]-\left[\begin{array}{rr}8 & 0 \\ -8 & 56\end{array}\right]-\left[\begin{array}{ll}\mathrm{k} & 0 \\ 0 & \mathrm{k}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]} \\ \therefore & {\left[\begin{array}{rc}1-8-\mathrm{k} & 0-0-0 \\ -8+8-0 & 49-56-\mathrm{k}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]}\end{array}$

∴ by equality of matrices, we get 1 – 8 – k = 0 ∴ k = -7

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Question 333 Marks
Verify that A(B + C) = AB + AC in each of the following matrices: $A=\left[\begin{array}{ccc}1 & -1 & 3 \\ 2 & 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 0 \\ -2 & 3 \\ 4 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 2 \\ -2 & 0 \\ 4 & -3\end{array}\right]$
Answer
$A(B+C)=\left[\begin{array}{ccc}1 & -1 & 3 \\ 2 & 3 & 2\end{array}\right]\left\{\left[\begin{array}{cc}1 & 0 \\ -2 & 3 \\ 4 & 3\end{array}\right]+\left[\begin{array}{cc}1 & 2 \\ -2 & 0 \\ 4 & -3\end{array}\right]\right\}$

$=\left[\begin{array}{ccc}1 & -1 & 3 \\ 2 & 3 & 2\end{array}\right]\left[\begin{array}{cc}1+1 & 0+2 \\ -2-2 & 3+0 \\ 4+4 & 3-3\end{array}\right]$

$=\left[\begin{array}{ccc}1 & -1 & 3 \\ 2 & 3 & 2\end{array}\right]\left[\begin{array}{cc}2 & 2 \\ -4 & 3 \\ 8 & 0\end{array}\right]$

$=\left[\begin{array}{cc}2+4+24 & 2-3+0 \\ 4-12+16 & 4+9+0\end{array}\right]$

$=\left[\begin{array}{cc}30 & -1 \\ 8 & 13\end{array}\right]$

$\ldots$ (i)

$\begin{aligned} & \mathrm{AB}+\mathrm{AC} \\ & =\left[\begin{array}{ccc}1 & -1 & 3 \\ 2 & 3 & 2\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & 3 \\ 4 & 3\end{array}\right]+\left[\begin{array}{ccc}1 & -1 & 3 \\ 2 & 3 & 2\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -2 & 0 \\ 4 & -3\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}1+2+12 & 0-3+9 \\ 2-6+8 & 0+9+6\end{array}\right]+\left[\begin{array}{ll}1+2+12 & 2+0-9 \\ 2-6+8 & 4+0-6\end{array}\right] \\ & =\left[\begin{array}{cc}15 & 6 \\ 4 & 15\end{array}\right]+\left[\begin{array}{cc}15 & -7 \\ 4 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}15+15 & 6-7 \\ 4+4 & 15-2\end{array}\right]\end{aligned}$

$=\left[\begin{array}{cc}30 & -1 \\ 8 & 13\end{array}\right]$

$\ldots$..(ii)

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Question 343 Marks
Verify that A(B + C) = AB + AC in each of the following matrices: $A=\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right], B=\left[\begin{array}{cc}-1 & 1 \\ 3 & -2\end{array}\right]$ and $C==\left[\begin{array}{cc}4 & 1 \\ 2 & -1\end{array}\right]$
Answer
$\mathrm{A}(\mathrm{B}+\mathrm{C})=\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right]\left\{\left[\begin{array}{cc}-1 & 1 \\ 3 & -2\end{array}\right]+\left[\begin{array}{cc}4 & 1 \\ 2 & -1\end{array}\right]\right\}$

$\begin{aligned} & =\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}-1+4 & 1+1 \\ 3+2 & -2-1\end{array}\right] \\ & =\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}3 & 2 \\ 5 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}12-10 & 8+6 \\ 6+15 & 4-9\end{array}\right]\end{aligned}$

$=\left[\begin{array}{cc}2 & 14 \\ 21 & -5\end{array}\right]$

$\ldots(\mathrm{i})$

$\begin{aligned} & \mathrm{AB}+\mathrm{AC} \\ & =\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 3 & -2\end{array}\right]+\left[\begin{array}{cc}4 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}4 & 1 \\ 2 & -1\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{ll}-4-6 & 4+4 \\ -2+9 & 2-6\end{array}\right]+\left[\begin{array}{cc}16-4 & 4+2 \\ 8+6 & 2-3\end{array}\right] \\ & =\left[\begin{array}{cc}-10 & 8 \\ 7 & -4\end{array}\right]+\left[\begin{array}{cc}12 & 6 \\ 14 & -1\end{array}\right] \\ & =\left[\begin{array}{cc}-10+12 & 8+6 \\ 7+14 & -4-1\end{array}\right]\end{aligned}$

$=\left[\begin{array}{cc}2 & 14 \\ 21 & -5\end{array}\right]$

...(ii)

From (i) and (ii), we get A(B + C) = AB + AC. [Note: The question has been modified.]

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Question 353 Marks
Show that AB = BA, where

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right], B=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

Answer
$A B=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

$=\left[\begin{array}{cc}\cos ^2 \theta+\sin ^2 \theta & -\sin \theta \cos \theta+\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \sin ^2 \theta+\cos ^2 \theta\end{array}\right]$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\ldots$ (i)

$\begin{aligned} & \mathbf{B A}=\left[\begin{array}{cr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \\ & =\left[\begin{array}{cc}\cos ^2 \theta+\sin ^2 \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\ \sin \theta \cos \theta-\sin \theta \cos \theta & \sin ^2 \theta+\cos ^2 \theta\end{array}\right]\end{aligned}$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

...(ii)

From (i) and (ii), we get AB = BA [Note: The question has been modified.]

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Question 363 Marks
Show that AB = BA, where

$A=\left[\begin{array}{lll}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right], B=\left[\begin{array}{lll}1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1\end{array}\right]$

Answer
$\mathbf{A B}=\left[\begin{array}{ccc}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right]\left[\begin{array}{lll}1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1\end{array}\right]$

$=\left[\begin{array}{ccc}-2+6-3 & -6+6-0 & 2-3+1 \\ -1+4-3 & -3+4-0 & 1-2+1 \\ -6+18-12 & -18+18+0 & 6-9+4\end{array}\right]$

$\therefore \quad \mathrm{AB}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\ldots(\mathrm{i})$

$\begin{aligned} \mathbf{B A} & =\left[\begin{array}{lll}1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1\end{array}\right]\left[\begin{array}{lll}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right] \\ & =\left[\begin{array}{lll}-2-3+6 & 3+6-9 & -1-3+4 \\ -4-2+6 & 6+4-9 & -2-2+4 \\ -6+0+6 & 9+0-9 & -3+0+4\end{array}\right]\end{aligned}$

$\therefore \quad B A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\ldots$ (ii)

From (i) and (ii), we get AB = BA

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Question 373 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
$A B=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & -3 & 1\end{array}\right]\left[\begin{array}{lll}2 & 1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}-2+3+1 & -1+0+2 & -4+2+1 \\ 4+9+0 & 2+0+0 & 8+6+0 \\ 2-9+1 & 1+0+2 & 4-6+1\end{array}\right]$

$\therefore \quad \mathrm{AB}=\left[\begin{array}{ccc}2 & 1 & -1 \\ 13 & 2 & 14 \\ -6 & 3 & -1\end{array}\right]$

...(i)

$\mathrm{BA}=\left[\begin{array}{lll}2 & 1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & -3 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}-2+2+4 & 2+3-12 & 2+0+4 \\ -3+0+2 & 3+0-6 & 3+0+2 \\ -1+4+1 & 1+6-3 & 1+0+1\end{array}\right]$

$\therefore \quad B A=\left[\begin{array}{ccc}4 & -7 & 6 \\ -1 & -3 & 5 \\ 4 & 4 & 2\end{array}\right]$

$\ldots$ (ii)

From (i) and (ii), we get AB ≠ BA

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Question 383 Marks
If $\left[\begin{array}{cc}2 a+b & 3 a-b \\ c+2 d & 2 c-d\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & -1\end{array}\right]$, find $\mathrm{a}, \mathrm{b}, c$ and $d$
Answer
$\left[\begin{array}{cc}2 a+b & 3 a-b \\ c+2 d & 2 c-d\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & -1\end{array}\right]$

∴ By equality of matrices, we get 2a + b = 2 ….(i) 3a – b = 3 ….(ii) c + 2d = 4 ….(iii) 2c – d = -1 ….(iv) Adding (i) and (ii), we get 5a = 5 ∴ a = 1 Substituting a = 1 in (i), we get 2(1) + b = 2 ∴ b = 0 By (iii) + (iv) x 2, we get 5c = 2

$\therefore c=\frac{2}{5}$

Substituting $c=\frac{2}{5}$ in (iii), we get

$\begin{aligned} & \frac{2}{5}+2 d=4 \\ & \therefore 2 d=4-\frac{2}{5} \\ & \therefore 2 d=\frac{18}{5} \\ & \therefore d=\frac{9}{5}\end{aligned}$

[Note: Answer given in the textbook is $d=\frac{3}{5}$.

However, as per our calculation it is $d=\frac{9}{5}$.]

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Question 393 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
$\left[\begin{array}{ccc}2 x+y & -1 & 1 \\ 3 & 4 y & 4\end{array}\right]+\left[\begin{array}{ccc}-1 & 6 & 4 \\ 3 & 0 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & 5 & 5 \\ 6 & 18 & 7\end{array}\right]$

$\therefore \quad\left[\begin{array}{ccc}2 x+y-1 & -1+6 & 1+4 \\ 3+3 & 4 y+0 & 4+3\end{array}\right]=\left[\begin{array}{ccc}3 & 5 & 5 \\ 6 & 18 & 7\end{array}\right]$

$\therefore \quad\left[\begin{array}{ccc}2 x+y-1 & 5 & 5 \\ 6 & 4 y & 7\end{array}\right]=\left[\begin{array}{ccc}3 & 5 & 5 \\ 6 & 18 & 7\end{array}\right]$

∴ By equality of matrices, we get 2x + y – 1 = 3 and 4y = 18

$\begin{aligned} & \therefore 2 x+y=4 \text { and } y=\frac{18}{4}=\frac{9}{2} \\ & \therefore 2 x+\frac{9}{2}=4 \\ & \therefore 2 x=4-\frac{9}{2} \\ & \therefore 2 x=\frac{1}{2}= \\ & \therefore x=-\frac{1}{4}=\text { and } y=\frac{9}{2}=\end{aligned}$

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Question 403 Marks
If $\mathrm{A}=\left[\begin{array}{cc}1 & 2 i \\ -3 & 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}2 i & 1 \\ 2 & -3\end{array}\right]$ where $i=\sqrt{-1}$, find $\mathrm{A}+\mathrm{B}$ and $\mathrm{A}-\mathrm{B}$. Show that $\mathrm{A}$

+ B is singular. Is A – B singular? Justify your answer.

Answer
$\begin{aligned} A+B & =\left[\begin{array}{cc}i & 2 i \\ -3 & 2\end{array}\right]+\left[\begin{array}{cc}2 i & i \\ 2 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}i+2 i & 2 i+i \\ -3+2 & 2-3\end{array}\right]=\left[\begin{array}{cc}3 i & 3 i \\ -1 & -1\end{array}\right]\end{aligned}$

$\therefore \quad|A+B|=\left|\begin{array}{cc}3 \mathrm{i} & 3 \mathrm{i} \\ -1 & -1\end{array}\right|$

$=3 \mathrm{i}(-1)-(-1) 3 \mathrm{i}=0$

$\therefore \quad \mathrm{A}+\mathrm{B}$ is a singular matrix.

$\begin{aligned} A-B & =\left[\begin{array}{cc}i & 2 i \\ -3 & 2\end{array}\right]-\left[\begin{array}{cc}2 i & i \\ 2 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}i-2 i & 2 i-i \\ -3-2 & 2+3\end{array}\right]=\left[\begin{array}{cc}-i & i \\ -5 & 5\end{array}\right]\end{aligned}$

$\therefore \quad|\mathrm{A}-\mathrm{B}|=\left|\begin{array}{ll}-\mathrm{i} & \mathrm{i} \\ -5 & 5\end{array}\right|=(-\mathrm{i}) 5-(-5) \mathrm{i}=0$

$\therefore \quad \mathrm{A}-\mathrm{B}$ is also a singular matrix.

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Question 413 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

3A – 4B + 5X = C.

Answer
3A-4B + 5X = C ∴ 5X = C + 4B – 3A

$=\left[\begin{array}{cc}2 & 4 \\ -1 & -4 \\ -3 & 6\end{array}\right]+4\left[\begin{array}{cc}-1 & -2 \\ 4 & 2 \\ 1 & 5\end{array}\right]-3\left[\begin{array}{cc}1 & -2 \\ 3 & -5 \\ -6 & 0\end{array}\right]$

$=\left[\begin{array}{cc}2 & 4 \\ -1 & -4 \\ -3 & 6\end{array}\right]+\left[\begin{array}{cc}-4 & -8 \\ 16 & 8 \\ 4 & 20\end{array}\right]-\left[\begin{array}{cc}3 & -6 \\ 9 & -15 \\ -18 & 0\end{array}\right]$

$=\left[\begin{array}{cc}2-4-3 & 4-8+6 \\ -1+16-9 & -4+8+15 \\ -3+4+18 & 6+20-0\end{array}\right]$

$\therefore \quad 5 X=\left[\begin{array}{cc}-5 & 2 \\ 6 & 19 \\ 19 & 26\end{array}\right]$

$\therefore \quad X=\frac{1}{5}\left[\begin{array}{cc}-5 & 2 \\ 6 & 19 \\ 19 & 26\end{array}\right]=\left[\begin{array}{cc}-1 & \frac{2}{5} \\ \frac{6}{5} & \frac{19}{5} \\ \frac{19}{5} & \frac{26}{5}\end{array}\right]$

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Question 423 Marks
Find the value of $k$, if the area of triangle whose vertices are $P(k, 0), Q(2,2), R(4,3)$ is $\frac{3}{2}$ sq.units.
Answer
Here, $P\left(x_1, y_1\right) \equiv P(k, 0), Q\left(x_2, y_2\right) \equiv Q(2,2), R\left(x_3, y_3\right) \equiv R(4,3)$

$\therefore \mathrm{A}(\triangle \mathrm{PQR})=\frac{3}{2}$ sq. units

Area if triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\therefore \pm \frac{3}{2}=\frac{1}{2}\left|\begin{array}{lll}k & 0 & 1 \\ 2 & 2 & 1 \\ 4 & 3 & 1\end{array}\right|$

$\therefore \pm[k(2-3)-0+1(6-8)] \therefore \pm\left[\right.$ latex $\frac{3}{2}=\frac{1}{2}(-k-2)$

∴ ± 3 = -k – 2 ∴ 3 = -k – 2 or -3 = -k – 2 ∴ k = -5 or k = 1

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Question 433 Marks
Find the area of triangle whose vertices are : M (0, 5), N (- 2, 3), T (1, – 4)
Answer
Here, $M\left(x_1, y_1\right) \equiv M(0,5), N\left(x_2, y_2\right) \equiv N(-2,3)$

$T\left(x_3, y_3\right) \equiv T(1,-4)$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\therefore A(\Delta M N T)=\frac{1}{2}\left|\begin{array}{ccc}0 & 5 & 1 \\ -2 & 3 & 1 \\ 1 & -4 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[0-5(-2-1)+1(8-3)] \\ & =\frac{1}{2}[-5(-3)+1(5)] \\ & =\frac{1}{2}(15+5) \\ & =\frac{1}{2}(20) \\ & =10 \text { sq. units }\end{aligned}$

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Question 443 Marks
Find the area of triangle whose vertices are : P(3/2, 1), Q(4,2), R(4, -1/2)
Answer
Here, $P\left(x_1, y_1\right) \equiv P(3 / 2,1), Q\left(x_2, y_2\right) \equiv Q(4,2), R\left(x_3, y_3\right) \equiv R\left(4,-\frac{1}{2}\right)$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\therefore \quad A(\Delta P Q R)=\frac{1}{2}\left|\begin{array}{ccc}\frac{3}{2} & 1 & 1 \\ 4 & 2 & 1 \\ 4 & \frac{-1}{2} & 1\end{array}\right|$

$=\frac{1}{2}\left[\frac{3}{2}\left(2+\frac{1}{2}\right)-1(4-4)+1(-2-8)\right]$

$\begin{aligned} & =\frac{1}{2}\left[\frac{3}{2}\left(\frac{5}{2}\right)-1(0)+1(-10)\right] \\ & =\frac{1}{2}\left(\frac{15}{4}-10\right) \\ & =\frac{1}{2}\left(\frac{-25}{4}\right) \\ & =\frac{-25}{8}\end{aligned}$

Since area cannot be negative A(ΔPQR) = 25/8 sq. units

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Question 453 Marks
Find the area of triangle whose vertices are : A (5,8), B (5,0), C (1,0)
Answer
Here, $A\left(x_1, y_1\right) \equiv A(5,8), B\left(x_2, y_2\right)=B(5,0), C\left(x_3, y_3\right)=C(1,0)$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\mathrm{A}(\triangle \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{lll}5 & 8 & 1 \\ 5 & 0 & 1 \\ 1 & 0 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[5(0-0)-8(5-1)+1(0-0)] \\ & =\frac{1}{2}[0-8(4)+0] \\ & =\frac{1}{2}(-32) \\ & =-16\end{aligned}$

Since area cannot be negative, A(ΔABC) = 16 sq. units

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Question 463 Marks
Without expanding determinant, show that

$\left|\begin{array}{lll}1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12\end{array}\right|+4\left|\begin{array}{lll}2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6\end{array}\right|=10\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6\end{array}\right|$

Answer
L.H.S. $=\left|\begin{array}{ccc}1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12\end{array}\right|+4\left|\begin{array}{lll}2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6\end{array}\right|$

In 1st determinant, taking 2 common from $C_3$ we get

L.H.S. $=\left|\begin{array}{ccc}1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12\end{array}\right|+4\left|\begin{array}{lll}2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6\end{array}\right|$

In $1^{\text {st }}$ determinant, taking 2 common from $\mathrm{C}_3$,

we get

L.H.S. $=2\left|\begin{array}{lll}1 & 3 & 3 \\ 6 & 1 & 2 \\ 3 & 7 & 6\end{array}\right|+4\left|\begin{array}{lll}2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6\end{array}\right|$

$=\left|\begin{array}{lll}2 & 3 & 3 \\ 12 & 1 & 2 \\ 6 & 7 & 6\end{array}\right|+\left|\begin{array}{lll}8 & 3 & 3 \\ 8 & 1 & 2 \\ 4 & 7 & 6\end{array}\right|$

$=\left|\begin{array}{ccc}2+8 & 3 & 3 \\ 12+8 & 1 & 2 \\ 6+4 & 7 & 6\end{array}\right|$

$=\left|\begin{array}{lll}10 & 3 & 3 \\ 20 & 1 & 2 \\ 10 & 7 & 6\end{array}\right|$

Interchanging rows and columns, we get

L.H.S. $=\left|\begin{array}{ccc}10 & 20 & 10 \\ 3 & 1 & 7 \\ 3 & 2 & 6\end{array}\right|$

Taking 10 common from $R_1$, we get

L. H.S $=10\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6\end{array}\right|=$ R.H.S.

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Question 473 Marks
If $\left|\begin{array}{ccc}4+x & 4-x & 4-x \\ 4-x & 4+x & 4-x \\ 4-x & 4-x & 4+x\end{array}\right|=0$, then find the values of $x$.
Answer
$\left|\begin{array}{lll}4+x & 4-x & 4-x \\ 4-x & 4+x & 4-x \\ 4-x & 4-x & 4+x\end{array}\right|=0$

Applying $C_1 \rightarrow C_1+C_2+C_3$, we get

$\left|\begin{array}{lll}12-x & 4-x & 4-x \\ 12-x & 4+x & 4-x \\ 12-x & 4-x & 4+x\end{array}\right|=0$

Taking $(12-x)$ common from $C_1$, we get

$(12-x)\left|\begin{array}{ccc}1 & 4-x & 4-x \\ 1 & 4+x & 4-x \\ 1 & 4-x & 4+x\end{array}\right|=0$

Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$,

we get

$(12-x)\left|\begin{array}{ccc}1 & 4-x & 4-x \\ 0 & 2 x & 0 \\ 0 & 0 & 2 x\end{array}\right|=0$

$\begin{aligned} & (12-x)\left[1\left(4 x^2-0\right)-(4-x)(0-0)+(4-x)(0-0)\right]=0 \\ & \therefore(12-x)\left(4 x^2\right)=0 \\ & \therefore x^2(12-x)=0 \\ & \therefore x=0 \text { or } 12-x=0 \\ & \therefore x=0 \text { or } x=12\end{aligned}$

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Question 483 Marks
Prove that $\left|\begin{array}{lll}x+y & y+\mathbf{z} & \mathbf{z}+x \\ \mathbf{z}+x & x+y & y+\mathbf{z} \\ y+\mathbf{z} & \mathbf{z}+x & x+y\end{array}\right|=2\left|\begin{array}{lll}x & y & \mathbf{z} \\ \mathbf{z} & x & y \\ y & \mathbf{z} & x\end{array}\right|$
Answer
L.H.S. $=\left|\begin{array}{lll}x+y & y+z & z+x \\ z+x & x+y & y+z \\ y+z & z+x & x+y\end{array}\right|$

Applying $R_1 \rightarrow R_1+R_2+R_3$, we get

L.H.S.

$=\left|\begin{array}{ccc}2(x+y+z) & 2(x+y+z) & 2(x+y+z) \\ \mathrm{z}+x & x+y & y+z \\ y+z & z+x & x+y\end{array}\right|$

Taking 2 common from $R_1$, we get

L.H.S. $=2\left|\begin{array}{ccc}x+y+z & x+y+z & x+y+z \\ z+x & x+y & y+z \\ y+z & \mathrm{z}+x & x+y\end{array}\right|$

Applying $R_1 \rightarrow R_1-R_3$, we get

L.H.S. $=2\left|\begin{array}{ccc}x & y & \mathrm{z} \\ \mathrm{z}+x & x+y & y+\mathrm{z} \\ y+\mathrm{z} & \mathrm{z}+x & x+y\end{array}\right|$

Applying $R_3 \rightarrow R_3-R_2$, we get

L.H.S. $=2\left|\begin{array}{lll}x & y & z \\ z & x & y \\ y & z & x\end{array}\right|=$ RHS

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Question 493 Marks
Find the value of determinant expanding along third column $\left|\begin{array}{ccc}-1 & 1 & 2 \\ -2 & 3 & -4 \\ -3 & 4 & 0\end{array}\right|$
Answer
Here, $\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|=\left|\begin{array}{ccc}-1 & 1 & 2 \\ -2 & 3 & -4 \\ -3 & 4 & 0\end{array}\right|$
Expantion along the third column
$=a_{13} C_{13}+a_{23} C_{23}+a_{33} C_{33}$
$=2 \times(-1)^{1+3}\left|\begin{array}{ll}-2 & 3 \\ -3 & 4\end{array}\right|-4 \times(-1)^{2+3}\left|\begin{array}{ll}-1 & 1 \\ -3 & 4\end{array}\right|+0 \times(-1)^{3+3}\left|\begin{array}{ll}-1 & 1 \\ -2 & 3\end{array}\right|$
$ =2(-8+9)+4(-4+3)+0$
$ =2-4$
$ =-2$
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Question 503 Marks
Find $x$ and $y$ if $\left|\begin{array}{ccc}4 i & i ^3 & 2 i \\ 1 & 3 i^2 & 4 \\ 5 & -3 & i\end{array}\right|= x + i y$, where $i ^2=-1$
Answer
$\begin{aligned}
& \left|\begin{array}{ccc}
4 i & i ^3 & 2 i \\
1 & 3 i ^2 & 4 \\
5 & -3 & i
\end{array}\right| \\
= & \left|\begin{array}{ccc}
4 i & - i & 2 i \\
1 & -3 & 4 \\
5 & -3 & i
\end{array}\right| \quad \quad \ldots\left[\because i ^2=-1\right] \\
= & 4 i \left|\begin{array}{ll}
-3 & 4 \\
-3 & i
\end{array}\right|-(- i )\left|\begin{array}{ll}
1 & 4 \\
5 & i
\end{array}\right|+2 i \left|\begin{array}{ll}
1 & -3 \\
5 & -3
\end{array}\right| \\
= & 4 i(-3 i +12)+ i ( i -20)+2 i(-3+15) \\
= & 12 i ^2+48 i + i ^2-20 i +24 i \\
= & -11 i ^2+52 i \\
= & -11(-1)+52 i \ldots\left[\because i ^2=-1\right] \\
= & 11+52 i
\end{aligned}
$
Comparing with $x+i y$, we get $x=11, y=52$
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