If A= 2&5&9 6&1&3 4&8&2 , find |A| - Vidyadip

MCQ

If $\text{A}=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix},$ find $|\text{A}|$

✓
$352$
B
$356$
C
$325$
D
$532$

✓

Answer

Correct option: A.

$352$

$\Rightarrow|\text{A}|=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}$ Evaluating along the first row, we get
$\triangle=2\begin{bmatrix}1&3\\8&2\end{bmatrix}-5\begin{bmatrix}6&3\\4&2\end{bmatrix}+9\begin{bmatrix}6&1\\4&8\end{bmatrix}$
$\triangle=2(2-24)-5(12-12)+9(48-4)$
$\triangle=2(-22)-0+9(44) $
$\triangle=-44+9(44)=44(-1+9)=352 $

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 "M.C.Q (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

let $S = \{1, 2, … 20\}$. A subset $B$ of $S$ is said to be $“nice”$, if the sum of the elements of $B$ is $203$. Then the probability that a randomly chosen subset of $S$ is $‘nice’$ is

The matrix $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a :

The corner points of the feasible region determined by the following system of linear inequalities:
$2\text{x}+\text{y}\le10,\ \text{x}+3\text{y}\le15,\ \text{x},\ \text{y}\ge0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is:

Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability that both cards are queen is

$\int_{}^{} {{{\cos }^5}x\;dx = } $

The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4 , respectively and $\vec{a} \cdot \vec{b}=2 \sqrt{3}$ is

$\int_{}^{} {\frac{{x + \sin x}}{{1 + \cos x}}\;dx} $ is equal to

Let A and B be two events such that P(A) = 0.6, P(B) = 0.2, P(A|B) = 0.5. Then $\text{P}(\overline{\text{A}}|\overline{\text{B}})$ equals.

If $\left[\begin{array}{cc}3 c+6 & a-d \\ a+d & 2-3 b\end{array}\right]=\left[\begin{array}{cc}12 & 2 \\ -8 & -4\end{array}\right]$ then find $a b-c d$ :

Let A = {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is: