If A = begin{bmatrix} x & 0 1 & 1 end{bmatri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $A = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix}$.First,calculate $A^2 = A \cdot A$:$A^2 = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} \be

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given $A = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix}$.First,calculate $A^2 = A \cdot A$:$A^2 = \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^2 + 0 & 0 + 0 \ x + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \ x + 1 & 1 \end{bmatrix}$.Next,calculate $A^3 = A^2 \cdot A$:$A^3 = \begin{bmatrix} x^2 & 0 \ x + 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^3 + 0 & 0 + 0 \ x(x + 1) + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^3 & 0 \ x^2 + x + 1 & 1 \end{bmatrix}$.Given $A^3 = B$,we have:$\begin{bmatrix} x^3 & 0 \ x^2 + x + 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 \ 7 & 1 \end{bmatrix}$.Comparing the corresponding elements:$x^3 = 8 \implies x = 2$.Also,$x^2 + x + 1 = 7 \implies x^2 + x - 6 = 0$.Factoring the quadratic equation: $(x + 3)(x - 2) = 0$,which gives $x = 2$ or $x = -3$.Since $x$ must satisfy both conditions,we take the common value,$x = 2$.

If $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$ and $A^3 = B$,then $x =$

Similar Questions

Find the value of $x, y$ and $z$ from the following equation : $\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$

If ${a_{ij}} = \frac{1}{2}(3i - 2j)$ and $A = {[{a_{ij}}]_{2 \times 2}}$,then $A$ is equal to

For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$,find the values of $a$ and $b$ such that $A^{2} + aA + bI = 0$.

If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,find $k$ such that $A^{2} = kA - 2I$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5 \end{bmatrix}$,then $AB$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module