Solve the Following Question.(2 Marks)

(ii) $\left|\begin{array}{lll}1 & y z & y+z \\ 1 & z x & z+x \\ 1 & x y & x+y\end{array}\right|=\left|\begin{array}{lll}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right|$

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Question 142 Marks

Without expanding determinants, prove that

(i) $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=\left|\begin{array}{lll}b_1 & c_1 & a_1 \\ b_2 & c_2 & a_2 \\ b_3 & c_3 & a_3\end{array}\right|=\left|\begin{array}{lll}c_1 & a_1 & b_1 \\ c_2 & a_2 & b_2 \\ c_3 & a_3 & b_3\end{array}\right|$

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Question 152 Marks

Without expanding determinants, find the value of

(ii) $\left|\begin{array}{lll}2014 & 2017 & 1 \\ 2020 & 2023 & 1 \\ 2023 & 2026 & 1\end{array}\right|$

Answer

Let $\mathrm{D}=\left|\begin{array}{lll}2014 & 2017 & 1 \\ 2020 & 2023 & 1 \\ 2023 & 2026 & 1\end{array}\right|$
Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1$, we get
$
D=\left|\begin{array}{lll}
2014 & 3 & 1 \\
2020 & 3 & 1 \\
2023 & 3 & 1
\end{array}\right|
$
Taking (3) common from $\mathrm{C}_2$, we get
$
\begin{aligned}
\mathrm{D} & =3\left|\begin{array}{lll}
2014 & 1 & 1 \\
2020 & 1 & 1 \\
2023 & 1 & 1
\end{array}\right| \text { } \\
& =3(0) \\
& =3
\end{aligned}
$

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Question 162 Marks

Without expanding determinants, find the value of

(i) $\left|\begin{array}{lll}10 & 57 & 107 \\ 12 & 64 & 124 \\ 15 & 78 & 153\end{array}\right|$

Answer

Let $D=\left|\begin{array}{lll}10 & 57 & 107 \\ 12 & 64 & 124 \\ 15 & 78 & 153\end{array}\right|$
Applying $\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2$, we get
$
\mathrm{D}=\left|\begin{array}{lll}
10 & 57 & 50 \\
12 & 64 & 60 \\
15 & 78 & 75
\end{array}\right|
$
Taking (5) common from $\mathrm{C}_3$, we get
$
\begin{aligned}
& \mathrm{D}=5\left|\begin{array}{lll}
10 & 57 & 10 \\
12 & 64 & 12 \\
15 & 78 & 15
\end{array}\right| \\
& =5(0) \quad \ldots\left[\because C_1 \text { and } \mathrm{C}_3 \text { are identical }\right] \\
& =0 \quad \text { } \\
&
\end{aligned}
$

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Question 172 Marks

Without expanding determinants, show that
$
\left|\begin{array}{ccc}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{ccc}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|=10\left|\begin{array}{ccc}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|
$

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Question 182 Marks

Find $\mathrm{x}$ and $\mathrm{y}$ if $\left|\begin{array}{ccc}4 i & i^3 & 2 i \\ 1 & 3 i^2 & 4 \\ 5 & -3 & i\end{array}\right|=\mathrm{x}+\mathrm{iy}$, where $\mathrm{i}=\sqrt{ }-1$

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Question 192 Marks

Find the value of $x$, if
$
\left|\begin{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
\end{array}\right|=29
$

Answer

\begin{aligned}
& \quad\left|\begin{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
\end{array}\right|=29 \\
& \therefore \quad x\left|\begin{array}{cc}
1 & -3 \\
-4 & 5
\end{array}\right|-(-1)\left|\begin{array}{cc}
2 x & -3 \\
3 & 5
\end{array}\right|+2\left|\begin{array}{cc}
2 x & 1 \\
3 & -4
\end{array}\right|=29 \\
& \therefore x(5-12)+1(10 x+9)+2(-8 x-3)=29 \\
& \therefore-7 x+10 x+9-16 x-6=29 \\
& \therefore-13 x+3=29 \\
& \therefore-13 x=26 \\
& \therefore x=-2
\end{aligned}

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Maths (commerce) Solve the Following Question.(2 Marks)

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