Let the determinant of a square matrix A of o | vedclass

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

Question

(A) Given the system of equations: $4m + n = 22$ $(1)$ $17m + 4n = 93$ $(2)$ Multiplying $(1)$ by $4$,we get $16m + 4n = 88$. Subtracting this from $(

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(A) Given the system of equations: $4m + n = 22$ $(1)$ $17m + 4n = 93$ $(2)$ Multiplying $(1)$ by $4$,we get $16m + 4n = 88$. Subtracting this from $(2)$,we get $m = 5$. Substituting $m = 5$ into $(1)$,$20 + n = 22$,so $n = 2$. Thus,the order of matrix $A$ is $m = 5$,and $|A| = m - n = 5 - 2 = 3$. We need to find $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(mA)))$. Since $n = 2$ and $m = 5$,this is $\operatorname{det}(2 \operatorname{adj}(\operatorname{adj}(5A)))$. Using the property $\operatorname{det}(kA) = k^m \operatorname{det}(A)$ for a matrix of order $m$: $\operatorname{det}(2 \operatorname{adj}(\operatorname{adj}(5A))) = 2^5 \operatorname{det}(\operatorname{adj}(\operatorname{adj}(5A)))$. Using $\operatorname{det}(\operatorname{adj}(B)) = |B|^{m-1}$,we have $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(5A))) = |\operatorname{adj}(5A)|^{5-1} = |\operatorname{adj}(5A)|^4$. Since $|\operatorname{adj}(5A)| = |5A|^{5-1} = |5A|^4 = (5^5 |A|)^4 = 5^{20} |A|^4$. So,$\operatorname{det}(\operatorname{adj}(\operatorname{adj}(5A))) = (5^{20} |A|^4)^4 = 5^{80} |A|^{16}$. Thus,$\operatorname{det}(2 \operatorname{adj}(\operatorname{adj}(5A))) = 2^5 \cdot 5^{80} \cdot 3^{16}$. Since $6^5 = 2^5 \cdot 3^5$,we rewrite the expression: $2^5 \cdot 5^{80} \cdot 3^{16} = 3^{11} \cdot 5^{80} \cdot (2^5 \cdot 3^5) = 3^{11} \cdot 5^{80} \cdot 6^5$. Comparing with $3^a 5^b 6^c$,we get $a = 11, b = 80, c = 5$. Therefore,$a + b + c = 11 + 80 + 5 = 96$.

Let the determinant of a square matrix $A$ of order $m$ be $m-n$,where $m$ and $n$ satisfy $4m + n = 22$ and $17m + 4n = 93$. If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(mA))) = 3^a 5^b 6^c$,then $a + b + c$ is equal to:

Similar Questions

If $\text{adj} \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{bmatrix}$, then $m+n=$

Let $A$ be an invertible square matrix of order $3 \times 3$. Then $|(\text{adj} A) \cdot A|$ is

Let $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI_2$,(where $I_2$ is the unit matrix of order $2$),then

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

Let $A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix}$. If $AA^T = I_3$,then $|p|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module