If A= 1&2 3&-1 and B= 1&-4 3&-2 , find |AB|. - Vidyadip

Question

If $\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix},$ find |AB|.

✓

Answer

$\Rightarrow\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$
⇒ |A| = -1 - 6 = -7
$\Rightarrow\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
⇒ |B| = -2 + 12 = 10
If A and B are square matrix of the same order, then |AB| = |A| |B|.
⇒ |AB| = |A| |B|
⇒ |AB| = -7 × 10 = -70

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 "Solve the Following Question.(2 Marks)" from Determinants→Determinants — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

$\int_0^4 \frac{1}{\sqrt{4 x-x^2}} \cdot d x$

Find the area of the region bounded by the line $2 y+x=8$, $x$-axis and the lines $x=2$ and $x=4$.

Define an equivalence relation.

If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}$, then show that $\frac{d y}{d x}=\frac{1}{x(2 y-1)}$

Evaluate the following:
$\cot\Big(\cot^{-1}\frac{3}{5}\Big)$

If $A=\left[\begin{array}{ll}6 & 5 \\ 5 & 6\end{array}\right]$ and $B=\left[\begin{array}{cc}11 & 0 \\ 0 & 11\end{array}\right]$ then find $A^{\prime} B^{\prime}$

Compute the products AB and BA whichever exists the following cases:
$[\text{a},\text{b}]\begin{bmatrix}\text{c}\\\text{d} \end{bmatrix}+\big[\text{a},\text{b},\text{c},\text{d}\big]\begin{bmatrix}\text{a}\\\text{b}\\\text{c}\\\text{d}\end{bmatrix}$

$\int \frac{\sin (x-a)}{\cos (x+ b )} d x$

Determine whether the following operations define a binary operation on the given set or not:$'\odot'$ on N defined by $\text{a}\odot\text{b}=\text{a}^{\text{b}}+\text{b}^{\text{a}}$ for all $\text{a, b}\in\text{N.}$

If the tangent to a curve at a point (x, y) is equally inclined to the co-ordinates axes then write the value of $\frac{\text{dy}}{\text{dx}}.$