(C) Given that $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$.Multiplying matrix $A$ by a scalar $h$, we get:$hA = \begin{bmatrix} 0 & 2h \\ 3h &

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) Given that $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$.Multiplying matrix $A$ by a scalar $h$, we get:$hA = \begin{bmatrix} 0 & 2h \\ 3h & -4h \end{bmatrix}$.We are given that $hA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$.Comparing the corresponding elements of the two matrices:$1$) $-4h = 24 \implies h = -6$.$2$) $2h = 3a \implies 2(-6) = 3a \implies -12 = 3a \implies a = -4$.$3$) $3h = 2b \implies 3(-6) = 2b \implies -18 = 2b \implies b = -9$.Thus, the values are $h = -6, a = -4, b = -9$.

Class 12 Mathematics 3-4. Matrices and Determinants Types of matrices, Algebra of matrices

Easy

If $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$ and $hA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$, then the values of $h, a, b$ are respectively

TS EAMCET 2001 All TS EAMCET

AP EAMCET 2001 All AP EAMCET

A
$-6, -12, -18$
B
$-6, 4, 9$
C
$-6, -4, -9$
D
$-6, 12, 18$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

3-4. Matrices and Determinants

Subtopic

Types of matrices, Algebra of matrices

Previous Year

TS EAMCET 2001 Paper All TS EAMCET years →

Previous Year

AP EAMCET 2001 Paper All AP EAMCET years →

If $P = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ -2 & 0 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{bmatrix}$,then $P_{22} = $

Medium

View Solution

If $m[-3, 4] + n[4, -3] = [10, -11]$, then $3m + 7n$ is equal to

EasyAP EAMCET 2005

View Solution

If $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$,then prove that $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$,where $n$ is any positive integer.

Medium

View Solution

Find the values of $x$ and $y$ from the following equation:
$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

Easy

View Solution

If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$,then $AB = BA$ (given $B \neq I$). Which of the following matrices $B$ satisfies this condition?

EasyGUJCET 2018

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$ and $hA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$, then the values of $h, a, b$ are respectively

Similar Questions

If $P = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ -2 & 0 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{bmatrix}$,then $P_{22} = $

If $m[-3, 4] + n[4, -3] = [10, -11]$, then $3m + 7n$ is equal to

If $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$,then prove that $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$,where $n$ is any positive integer.

Find the values of $x$ and $y$ from the following equation:$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$,then $AB = BA$ (given $B \neq I$). Which of the following matrices $B$ satisfies this condition?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the values of $x$ and $y$ from the following equation:
$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$