MCQ
If $A$ is a square matrix such that $A^2=A$, then $(1+A)^3-7 A$ is equal to:
  • A
    A
  • B
    I - A
  • I
  • D
    3A

Answer

Correct option: C.
I
Given: $A^2=A \ldots (i)$
Multiplying both sides by $A, A^3=A^2=A[$ From eq. (i) $] \ldots (ii)$
Also given $(I+A)^3-7 A=I^3+A^3+3 I^2 A+3 I A^2-7 A$
Putting $A^2=A\left[\right.$ from eq. (i)] and $A^3=A[$ from eq. $(ii)],$
$=I+A+3 I A+3 I A-7 A=I+A+3 A+3 A-7 A[\because I A=A]$
$=I+7 A-7 A=I$
Therefore, option $(C)$ is correct.

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

Similar questions

Consider the system of equations $\mathrm{ax}+\mathrm{by}=0, \mathrm{cx}+\mathrm{dy}=0$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in\{0,1\}$.

$STATEMENT -1$ : The probability that the system of equations has a unique solution is $3 / 8$. and $STATEMENT - 2$: The probability that the system of equations has a solution is $1$ .

If $\begin{bmatrix}\text{r}+4&\text{amp; 6}\\3&\text{amp; 3}\end{bmatrix}=\begin{bmatrix}{5}&\text{amp;}\text{ r}+5\\\text{r+2}&\text{amp; 4}\end{bmatrix}$ then $\text{r}=$
Range of ${\sin ^{ - 1\,}}\left( {\frac{{1 + {x^2}}}{{2 + {x^2}}}} \right)$ is 
$\int_0^\infty {\frac{{\log \,(1 + {x^2})}}{{1 + {x^2}}}} \,dx = $
An owner of a lodge plans an extension which contains not more than 50 rooms. At least 5 must be executive single rooms. The number of executive double rooms should be at least 3 times the number of executive single rooms. He charges ₹ 3000 for executive double room and ₹ 1800 for executive single room per day. Formulate the above problem as L.P.P. to maximize the profit.
The line $y = mx + 1$ is a tangent to the curve $y^2 = 4x,$ if the value of $m$ is:
If $A = \left[ {\begin{array}{*{20}{c}}\alpha &2\\2&\alpha \end{array}} \right]$ and $|{A^3}|$=125, then $\alpha = $
If a, b, c are distinct, then the value of x satisfying $\begin{vmatrix}0&\text{x}^2-\text{a}&\text{x}^3-\text{b}\\\text{x}^2+\text{a}&0&\text{x}^2+\text{c}\\\text{x}^4+\text{b}&\text{x}-\text{c}&0\end{vmatrix}=0$ is:
Solution of the differential equation, $\frac{{dy}}{{dx}} = \frac{{1 - 2y - 4x}}{{1 + y + 2x}}$ is
Evaluate the determinants

$\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$