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MHTCET 23 Apr 2025 Shift 2 Paper

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Question : 110 of 150

Marks: +1, -0

If

A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right], B=\operatorname{adj} A

C=5 A

\frac{|\operatorname{adj} B|}{|C|}=

2
4
1
5

Solution:

Concept:

For a

3\times 3

matrix

A

,

\operatorname{adj}(\operatorname{adj} A)=|A|A

.

Explanation:

Given

B=\operatorname{adj} A

. So

\operatorname{adj} B=\operatorname{adj}(\operatorname{adj} A)

.

Using the property,

\operatorname{adj} B=|A|A

.

Now compute

|A|

for

A=\begin{bmatrix}1&-1&1\\0&2&-3\\2&1&0\end{bmatrix}

|A|=1\begin{vmatrix}2&-3\\1&0\end{vmatrix}-(-1)\begin{vmatrix}0&-3\\2&0\end{vmatrix}+1\begin{vmatrix}0&2\\2&1\end{vmatrix}

|A|=1(0+3)+1(0+6)+1(0-4)=3+6-4=5

Therefore

\operatorname{adj} B=5A

.

It is given that

C=5A

.

Hence

\operatorname{adj} B=C

, so

|\operatorname{adj} B|=|C|

.

Thus

\frac{|\operatorname{adj} B|}{|C|}=1

Answer:

\dfrac{|\operatorname{adj} B|}{|C|}=1

, i.e. Option C.

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