2 Marks

Question 12 Marks

If $\left|\begin{array}{ll}2 & 3 \\ y & x\end{array}\right|=3,\left|\begin{array}{ll}x & y \\ 4 & 2\end{array}\right|=5$ then find value of $x$ and $y$.

Answer

Given : $\left|\begin{array}{ll}2 & 3 \\ y & x\end{array}\right|=3,\left|\begin{array}{ll}x & y \\ 4 & 2\end{array}\right|=5$
$
\begin{array}{l}
\therefore \quad \begin{array}{l}
2 x-3 y=3 .....(1) \\
2 x-4 y=5 .......(2)
\end{array}
\end{array}
$
Subtract equation (2) from equation (1)
$
y=-2
$
On putting the value of $y$ in equation (1)$
\begin{aligned}
& & 2 x-3 \times(-2) & =3 \\
\Rightarrow & & 2 x+6 & =3 \\
\Rightarrow & & x & =\frac{-3}{2}
\end{aligned}
$

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Question 22 Marks

Find the value of $x$ if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$.

Answer

$\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$$
\begin{aligned}
\text { LHS } & =\left|\begin{array}{cc}
2 & 4 \\
5 & 1
\end{array}\right|=2-20=-18 \\
\text { RHS } & =\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|=\left(2 x^2-24\right) \\
\therefore \quad \text { Hence } \quad-18 & =2 x^2-24 \text { or } 2 x^2=24-18=6 \\
\therefore \quad x^2 & =3 \Rightarrow x= \pm \sqrt{3} \quad \text {}
\end{aligned}
$

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Question 32 Marks

Find the value of the determinant $\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x-1\end{array}\right|$.

Answer

Determinant $\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x-1\end{array}\right|$$
\begin{array}{l}
=\left(x^2-x+1\right)(x-1)-(x+1)(x-1) \\
=(x-1)\left[x^2-x+1-x-1\right] \\
=(x-1)\left[x^2-2 x\right]=x(x-1)(x-2)
\end{array}
$

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Question 42 Marks

If point $A (m,-1), B (2,1)$ and $C (4,5)$ are collinear, then find value of $m$.

Answer

Three points are collinear if :$
\begin{array}{l}
\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=0 \\
\Rightarrow \quad\left|\begin{array}{ccc}
m & -1 & 1 \\
2 & 1 & 1 \\
4 & 5 & 1
\end{array}\right|=0 \\
\Rightarrow m(1-5)-(-1)(2-4)+(1)(10-4)=0 \\
\Rightarrow \quad-4 m-2+6=0 \\
\Rightarrow \quad-4 m+4=0 \\
\Rightarrow \quad 4 m=4 \Rightarrow m=1 \text {}
\end{array}
$

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Question 52 Marks

If $A =\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$, then write the value of $A .(\operatorname{adj} A )$.

Answer

$A_{11}=4, A_{12}=-1, A_{21}=-2, A_{22}=3$$
\begin{aligned}
\operatorname{adj} A & =\left[\begin{array}{cc}
4 & -1 \\
-2 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
4 & -2 \\
-1 & 3
\end{array}\right] \\
\text { A. }(\operatorname{adj} A) & =\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\left[\begin{array}{cc}
4 & -2 \\
-1 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
10 & 0 \\
0 & 10
\end{array}\right] \text {}
\end{aligned}
$

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Question 62 Marks

If $A$ has order 3 square matrix and $|A|=9$, then find value of $|2 \operatorname{adj} A |$.

Answer

$
\begin{array}{rlrl}
\mid 2 \text { adj } A \mid & =2^3|\operatorname{adj} A| & & {\left[\because|k A|=k^n|A|\right]} \\
& =8 \mid A^{3-1} & & {\left[|\operatorname{adj} A|=|A|^{n-1}\right]} \\
& =8(9)^{3-1} & & \\
& =8(9)^2 & & \\
& =8 \times 81=648 \text {}
\end{array}
$

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Question 72 Marks

If $A$ has order 3 invertible matrix and $A^2=2 A$, then find value of $|2 A|$.

Answer

Given: $A ^2=2 A$
$
\begin{array}{lll}
\Rightarrow & \left|A^2\right|=|2 A| & \\
\Rightarrow & |A|^2=2^3|A| & {\left[\because|-k A|=k^n(A)\right]} \\
\Rightarrow & |A|=0 \text { or }|A|=8
\end{array}
$
but A has non-singular matrix.$
\begin{aligned}
& \therefore & |A| & =8 \\
& \therefore & |2 A| & =|A|^2=8^2=64 \quad \text {}
\end{aligned}
$

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Question 82 Marks

For which value of $k$, square matrix $\left[\begin{array}{cc}k & 8 \\ 4 & 2 k\end{array}\right]$ is non-invertible matrix?

Answer

Suppose $A=\left|\begin{array}{cc}k & 8 \\ 4 & 2 k\end{array}\right|$
A is non-invertible, then $| A |=0$
$
\begin{aligned}
\Rightarrow & & \left|\begin{array}{cc}
k & 8 \\
4 & 2 k
\end{array}\right| & =0 \\
\Rightarrow & & 2 k^2-32 & =0 \\
\Rightarrow & & k^2 & =16 \\
\Rightarrow & & k & = \pm 4
\end{aligned}
$

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Question 92 Marks

Show that:$
\left|\begin{array}{ccc}
1 & a & b \\
-a & 1 & c \\
-b & -c & 1
\end{array}\right|=1+a^2+b^2+c^2
$

Answer

On expanding along $R _1$$
\begin{array}{l}
=1\left(1+c^2\right)-a(-a+b c)+b(a c+b) \\
=1+c^2+a^2-a b c+a b c+b^2 \\
=1+a^2+b^2+c^2
\end{array}
$
Hence proved.

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Maths 2 Marks

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