MCQ
Let $A =$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ , then
- A$A^2 - 4A - 5I_3 = 0$
- B$A^{-1} = \frac{1}{5} (A - 4I_3)$
- C$A^2$ is invertible
- ✓All of the above
✓
Answer
Correct option: D.
All of the above
d
$A^2 =$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $=$ $\left[ {\begin{array}{*{20}{c}}9&8&8\\8&9&8\\8&8&9\end{array}} \right]$
We have $A^2 - 4A - 5I_3$
$A^2 =$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $=$ $\left[ {\begin{array}{*{20}{c}}9&8&8\\8&9&8\\8&8&9\end{array}} \right]$
We have $A^2 - 4A - 5I_3$
$=$ $\left[ {\begin{array}{*{20}{c}}9&8&8\\8&9&8\\8&8&9\end{array}} \right]$ $- 4$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $- 5$ $\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ $= O$
==> $5I_3 = A^2 - 4A = A(A - 4I_3)$
==>$I_3 = A$ $\left[ {\frac{1}{5}(A - 4{I_3})} \right]$
==> $A^{-1} =$ $\frac{1}{5}(A - 4{I_3})$
Note that $|A| = 5$. Since $|A^3| = |A|^3 = 5^3 \ne 0, A^3$ is invertibleSimilarly, $A^2$ is invertible
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function $Z=3 x+9 y$ maximum?
→Let $\overrightarrow{ a }$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i}+\hat{j}, \hat{i}+\hat{k}$ and $\hat{i}-\hat{j}, \hat{j}-\hat{k}$. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$ then the ordered pair $(\theta,|\vec{a} \times \vec{b}|)$ is equal to
→$i\,.\,(j \times k) + j\,.\,(k \times i) + k\,.\,(i \times j) = $
→The degree of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} - \sqrt {\frac{{dy}}{{dx}} - 3} = x$ is
→Set of equations $a + b - 2c = 0,$ $2a - 3b + c = 0$ and $a - 5b + 4c = \alpha $ is consistent for $\alpha$ equal to
→The solution set of the equation ${\sin ^{ - 1}}x = 2{\tan ^{ - 1}}x$ is
→The area of the region bounded by the curve $y = x|x|$, $x-$ axis and the ordinates $x = 1,\,\,x = - 1$ is given by
→If $a + b + c = 0$, then the solution of the equation $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$ is
→The order of differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=1$ is:
→- one
- two
- four
- zero
A random variable X has the following probability distribution:
→| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is: