Download App

Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics2 Marks
MCQ
From the matrix equation $AB = AC$ we can conclude that $B = C$, provided
  • A
    A is singular.
  • ✓
    A is non-singular
  • C
    A is symmetric.
  • D
    A is square
✓

Answer

Correct option: B.
A is non-singular
(B) $AB = AC$
$\Rightarrow A ^{-1} AB = A ^{-1} AC$
$\Rightarrow IB = IC$
$\Rightarrow B = C$
$\therefore \quad$ For $B = C , A ^{-1}$ must exist
$\Rightarrow A$ is non-singular

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 "MCQ" from Matrics→Matrics — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

$\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} x+\cos ^{-1} \frac{1}{x}=$
→
If the inverse of the matrix $\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]$ does not exist, then the value of $\alpha$ is
→
Which of the given values of $X$ and $Y$ make the following pairs of matrices equal? $\begin{bmatrix}3\text{x}+7&5\\\text{y}+1&2-3\text{x}\end{bmatrix},\begin{bmatrix}0&\text{y}-2\\8&4\end{bmatrix}$
→
If $|\bar{a}+\bar{b}|>|\bar{a}-\bar{b}|$, then the angle between $\bar{a}$ and $\bar{b}$ is
→
The general value of $\theta$ satisfying $\sin ^2 \theta+\sin \theta=2$ is
→
If $O$ is the origin, $OP = 3$ with direction ratios proportional to $-1, 2, -2$ then the coordinates of $P$ are:
→
If $R$ is a relation on the set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ given by $xRy ⇔ y = 3x,$ then $R =$
→
The line $y = mx + 1$ is a tangent to the curve $y^2 = 4x$, if the value of m is:
→
If $M$ and $N$ are the midpoints of the sides $B C$ and $C D$ respectively of a parallelogram $A B C D$, then $\overline{ AM }+\overline{ AN }=$
→
If the position vectors of the vertices of a triangle be $6 \hat{i}+4 \hat{j}+5 \hat{k}, 4 \hat{i}+5 \hat{j}+6 \hat{k}$ and $5 \hat{ i }+6 \hat{ j }+4 \hat{ k }$, then the triangle is
→