Download App

STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
If $A = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]$, then
  • A
    ${A^2} = A$
  • B
    ${B^2} = B$
  • ✓
    $AB \ne BA$
  • D
    $AB = BA$
✓

Answer

Correct option: C.
$AB \ne BA$
c
(c) Since ${A^2} = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 4}&2\\{ - 3}&{ - 6}\end{array}} \right] \ne A$

${B^2} = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\4&9\end{array}} \right] \ne B$

Now $AB = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&6\\3&0\end{array}} \right]$

and $BA = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 7}&4\end{array}} \right]$

Obviously, $AB \ne BA$.

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 STD 12 - 3 and 4 . determinant and metrices→STD 12 - 3 and 4 . determinant and metrices — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $f(x)$ be a non-negative continous function such that the area bounded by the curve $y = f(x)$, $x -$ axis and the ordinates $x = \frac{\pi }{4}$, $x = \beta > \frac{\pi }{4}$ is $\left( {\beta \sin \beta + \frac{\pi }{4}\cos \beta + \sqrt 2 \beta } \right)$. Then $f\;\left( {\frac{\pi }{2}} \right)$ is
→
The function $\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$ is :
→
Choose the correct answer from the given four options : The points at which the tangents to the curve $y=x^3-12 x+18$ are parallel to $x-$ axis are :
→
The probability that a leap year will have $53$ fridays or $53$ Saturdays is.
→
The inverse of $\left[\begin{array}{cc}-4 & 3 \\ 7 & -5\end{array}\right]$ is
→
If $\mathrm{a}_{\mathrm{r}}=\cos \frac{2 \mathrm{r} \pi}{9}+i \sin \frac{2 \mathrm{r} \pi}{9}, \mathrm{r}=1,2,3, \ldots, i=\sqrt{-1}$ then the determinant $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to :
→
The corner points of the feasible region are $A(0, 0), B(16, 0), C(8, 16)$ and $D(0, 24)$. The minimum value of the objective function $z = 300x + 190y$ is $ .........:$
→
$N$  characters of information are held on magnetic tape, in batches of  $x$ characters each; the batch processing time is $\alpha + \beta {x^2}$ seconds; $\alpha $ and $\beta $ are constants. The optimal value of $ x$ for fast processing is
→
If the function $f : R \rightarrow R$ be such that $f(x) = x - [x],$ where $[x]$ denotes the greatest integer less than or equal to $x,$ then $f^{-1}(x)$ is:
→
If the matrix AB is zero, then:
→