Evaluate the determinant | ccc 3 & -1 & -2 0 & 0 & -1 3 & -5 & 0 … - Vidyadip

Question

Evaluate the determinant $\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

✓

Answer

Given: $\left| {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 2} \\ 0&0&{ - 1} \\ 3&{ - 5}&0 \end{array}} \right|$
Expanding along first row, $3\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\ { - 5}&0 \end{array}} \right| - \left( { - 1} \right)\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\ 3&0 \end{array}} \right| + \left( { - 2} \right)\left| {\begin{array}{*{20}{c}} 0&0 \\ 3&{ - 5} \end{array}} \right|$
= 3(0 - 5) + 1(0 - (-3)) - 2(0 - 0)
= -15 + 3 - 0 = -12

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "1 Marks" from Determinants→Determinants — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}2&7&65\\3&8&75\\5&9&86\end{vmatrix}=0$

If $3 \sin ^{-1} x+4 \cos ^{-1} x=2 \pi$, then find the value of $x$.

Evaluate the following:
$\cos^{-1}(\cos5)$

Determine if f defined by $f(x)=\left\{\begin{array}{c}x^2 \sin \frac{1}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$ is a continuous function?

By $f(x)=2 x$, defined a function $f : A \rightarrow B$ is one-one and onto both. If $A =\{1,2,3,4\}$ then find set B.

Find the value of $\int \frac{x d x}{\left(1-x^2\right)^{3 / 2}}$ :

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Z	3	2	1	0	-1
P(Z)	0.3	0.2	0.4	0.1	0.05

The value of $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$ is

Find the value of a if $\begin{bmatrix} \text{a} - \text{b}& 2\text{a} + \text{c}\\ 2\text{a} - \text{b} & 3\text{c} + \text{d} \\ \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 &13 \\ \end{bmatrix}$

Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is parallel to the line $\frac{\text{x} + 3 }{3} =\frac{4 - \text{y}}{5} = \frac{\text{z} + 8}{6}.$