Let A = begin{bmatrix} 2k - 1 & 1 & 1 0 & 2k | vedclass

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

Question

(A) Step $1$: Calculate $\det(A)$. Since $A$ is an upper triangular matrix, $\det(A) = (2k - 1)^3$.Step $2$: Calculate $\det(	ext{adj } A)$. We know

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Step $1$: Calculate $\det(A)$. Since $A$ is an upper triangular matrix, $\det(A) = (2k - 1)^3$.Step $2$: Calculate $\det(	ext{adj } A)$. We know $\det(	ext{adj } A) = (\det A)^{n-1}$. Here $n=3$, so $\det(	ext{adj } A) = ((2k - 1)^3)^2 = (2k - 1)^6$.Step $3$: Calculate $\det(B)$. $B$ is a skew-symmetric matrix of order $3 	imes 3$. The determinant of a skew-symmetric matrix of odd order is $0$. Thus, $\det(B) = 0$.Step $4$: Calculate $\det(	ext{adj } B)$. Since $\det(B) = 0$, $\det(	ext{adj } B) = (\det B)^{3-1} = 0^2 = 0$.Step $5$: Solve the equation $(2k - 1)^6 + 0 = 11^6$. This implies $(2k - 1)^6 = 11^6$, so $2k - 1 = 11$ or $2k - 1 = -11$.Step $6$: If $2k - 1 = 11$, then $2k = 12$, so $k = 6$. Then $k - 5 = 6 - 5 = 1$.Step $7$: If $2k - 1 = -11$, then $2k = -10$, so $k = -5$. Then $k - 5 = -5 - 5 = -10$ (not in options).Therefore, the value is $1$.

Similar Questions

The multiplicative inverse of matrix $\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$ is

The inverse of matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

Let $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.

If $A$ is a square matrix of order $3$,then consider the following statements.
$I$. If $|A|=0$,then $|\operatorname{Adj} A|=0$
$II$. If $|A| \neq 0$,then $|A^{-1}|=|A|^{-1}$
Which of the above statements is/are true?

Inverse of the matrix $\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module