If A = begin{bmatrix} 5a & -b 3 & 2 end{bmat | vedclass

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

Question

(D) Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$.We know that $A \operatorname{adj} A = |A| I$,where $I$ is the identity matrix.Given $A

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$.We know that $A \operatorname{adj} A = |A| I$,where $I$ is the identity matrix.Given $A \operatorname{adj} A = AA^{T}$,so $|A| I = AA^{T}$.$|A| = (5a)(2) - (-b)(3) = 10a + 3b$.Thus,$|A| I = \begin{bmatrix} 10a + 3b & 0 \ 0 & 10a + 3b \end{bmatrix}$.Now,$AA^{T} = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \ 15a - 2b & 9 + 4 \end{bmatrix}$.Equating the two matrices:$10a + 3b = 25a^2 + b^2$ (from diagonal elements) and $15a - 2b = 0$ (from off-diagonal elements).From $15a - 2b = 0$,we get $b = \frac{15a}{2}$.Also,$|A| I = AA^{T}$ implies $|A| = \det(A) = \det(A^T)$,so $10a + 3b = 13$.Substituting $b = \frac{15a}{2}$ into $10a + 3b = 13$:$10a + 3(\frac{15a}{2}) = 13 \implies 20a + 45a = 26 \implies 65a = 26 \implies a = \frac{26}{65} = \frac{2}{5}$.Then $b = \frac{15}{2} 	imes \frac{2}{5} = 3$.Therefore,$5a + b = 5(\frac{2}{5}) + 3 = 2 + 3 = 5$.

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \operatorname{adj} A = AA^{T}$,then $5a + b =$

Similar Questions

If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$,then $A^{-1}=$

The inverse of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$ is

If $A$ is an invertible matrix of order $n$, then the determinant of $\operatorname{adj} A$ is equal to :

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$,then the inverse of the matrix $A^3$ is

If $A = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix}$,then $(A^{-1})^3$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module