Solve the following Question.(1 Marks)

Question 11 Mark

Evaluate: $\left|\begin{array}{ccc}1 & -3 & 12 \\ 0 & 2 & -4 \\ 9 & 7 & 2\end{array}\right|$

Answer

$\begin{aligned} & \left|\begin{array}{ccc}1 & -3 & 12 \\ 0 & 2 & -4 \\ 9 & 7 & 2\end{array}\right| \text { } \\ & =1\left|\begin{array}{rr}2 & -4 \\ 7 & 2\end{array}\right|-(-3)\left|\begin{array}{rr}0 & -4 \\ 9 & 2\end{array}\right|+12\left|\begin{array}{ll}0 & 2 \\ 9 & 7\end{array}\right| \\ & =1(4+28)+3(0+36)+12(0-18) \\ & =1(32)+3(36)+12(-18) \\ & =32+108-216 \\ & =-76\end{aligned}$

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Question 21 Mark

Evaluate: $\left|\begin{array}{ccc}2 & -5 & 7 \\ 5 & 2 & 1 \\ 9 & 0 & 2\end{array}\right|$

Answer

$\begin{aligned} & \left|\begin{array}{ccc}2 & -5 & 7 \\ 5 & 2 & 1 \\ 9 & 0 & 2\end{array}\right|=2\left|\begin{array}{cc}2 & 1 \\ 0 & 2\end{array}\right|-(-5)\left|\begin{array}{ll}5 & 1 \\ 9 & 2\end{array}\right|+7\left|\begin{array}{ll}5 & 2 \\ 9 & 0\end{array}\right| \\ & =2(4-0)+5(10-9)+7(0-18) \\ & =2(4)+5(1)+7(-18) \\ & =8+5-126 \\ & =-113\end{aligned}$

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Question 31 Mark

Without expanding, evaluate the following determinants.

(iii) $\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$

Answer

Let $D=\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$
Applying $C_3 \rightarrow C_3-9 C_2$, we get
$
\begin{array}{rlrl}
\mathrm{D} & =\left|\begin{array}{lll}
2 & 7 & 2 \\
3 & 8 & 3 \\
5 & 9 & 5
\end{array}\right| & \text { MaharashtraBoardSolutions.in } \\
& =0 & & \ldots\left[\because C_1 \text { and } C_3 \text { are identical }\right]
\end{array}
$

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Question 41 Mark

Without expanding, evaluate the following determinants.

(ii) $\left|\begin{array}{ccc}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x\end{array}\right|$

Answer

Let $\mathrm{D}=\left|\begin{array}{ccc}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x\end{array}\right|$
Taking ( $3 x$ ) common from $\mathrm{R}_3$, we get
$
\begin{aligned}
& \mathrm{D}=3 x\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 8 \\
2 & 3 & 4
\end{array}\right| \\
& \text { MaharashtraBoardSolutions.in } \\
&=(3 x)(0) \quad \ldots\left[\because \mathrm{R}_1 \text { and } \mathrm{R}_3 \text { are identical }\right] \\
&=0
\end{aligned}
$

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Question 51 Mark

Without expanding, evaluate the following determinants.

(i) $\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$

Answer

Let $D=\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$
Applying $\mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_2$, we get
$D=\left|\begin{array}{lll}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{array}\right|$
Taking $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common from $\mathrm{C}_3$, we get
$
\begin{aligned}
& D=(a+b+c)\left|\begin{array}{lll}
1 & a & 1 \\
1 & b & 1 \\
1 & c & 1
\end{array}\right| \\
& \therefore \quad \mathrm{D}=(\mathrm{a}+\mathrm{b}+\mathrm{c})(0) \\
& \therefore \quad \mathrm{D}=0 \\
&
\end{aligned}
$
$\ldots\left[\because \mathrm{C}_1\right.$ and $\mathrm{C}_3$ are identical $]$

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Maths (commerce) Solve the following Question.(1 Marks)

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