Download App

Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices3 Marks
Question
If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] \& I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] ;$ Find k, So that $\mathrm{A}^2=\mathrm{kA}$ - 2l.
✓

Answer

A2 = A.A
$ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {9 - 8}&{ - 6 + 4} \\ {12 - 8}&{ - 8 + 4} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&{ - 4} \end{array}} \right]$
A2 = kA - 2I.
$\Rightarrow$$\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&{ - 4} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
$\Rightarrow$$\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&{ - 4} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3k}&{ - 2k} \\ {4k}&{ - 2k} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&2 \end{array}} \right]$
$\Rightarrow$$\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3k}&{ - 2k} \\ {4k}&{ - 2k} \end{array}} \right]$

$\Rightarrow$ 3k = 3
$\Rightarrow$ k = 1

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 "3 Marks" from Matrices→Matrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following integrals:
$\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}$
→
If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$
→
Find fog and gof if:

f(x) = x + 1, g(x) = 2x + 3

→
Represent the following families of curves by forming the corresponding differential equation:
$(\text{x}-\text{a})^2-\text{y}^2=1$
→
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = x2 - 3x + 2 on [-1, 2]
→
Write the value of $\lambda$ for which the lines $\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$ and $\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$ are perpendicular to each other.
→
If f(x) = |x - 2| write whether f(2) exists or not.
→
$\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
→
Evaluate the following integrals:
$\int\tan^32\text{x}\sec2\text{x dx}$
→
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
zx-plane
→