If A=≤ft[{array}{rrr}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4{array}] then A(I+{adj} A)=

Correct Answer is : [9−22010−33−212]≤ft[{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12{array}]903−210−22−312

≤ft[{array}{ccc}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 11{array}]

≤ft[{array}{ccc}8 & -2 & 2 \\ 0 & 9 & -3 \\ 3 & -2 & 10{array}]

≤ft[{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12{array}]

≤ft[{array}{ccc}3 & 2 & -2 \\ 0 & 10 & 3 \\ -3 & 2 & 12{array}]

Correct Answer is : [9−22010−33−212]≤ft[{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12{array}]903−210−22−312

Test Index

MHTCET 5 May 2025 Shift 2 Paper

Question : 137 of 150

Marks: +1, -0

A=\left[\begin{array}{rrr}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]

A(I+\operatorname{adj} A)=

$\left[\begin{array}{ccc}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 11\end{array}\right]$
$\left[\begin{array}{ccc}8 & -2 & 2 \\ 0 & 9 & -3 \\ 3 & -2 & 10\end{array}\right]$
$\left[\begin{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12\end{array}\right]$
$\left[\begin{array}{ccc}3 & 2 & -2 \\ 0 & 10 & 3 \\ -3 & 2 & 12\end{array}\right]$

Solution:

Concept:

Use the property

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

to simplify the given expression.

Explanation:

Expand the product:

A(I+\operatorname{adj} A) = AI + A(\operatorname{adj} A) = A + |A|I

Compute the determinant of

A

|A| = \begin{vmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{vmatrix}

|A| = 1(8-6) + 2(0+9) + 2(0-6)

|A| = 2 + 18 - 12 = 8

Therefore:

A(I+\operatorname{adj}A) = A + 8I

Add

8

to each diagonal entry of

A

A + 8I = \begin{bmatrix} 1+8 & -2 & 2 \\ 0 & 2+8 & -3 \\ 3 & -2 & 4+8 \end{bmatrix}

= \begin{bmatrix} 9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12 \end{bmatrix}

Answer:

Option C:

\begin{bmatrix} 9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12 \end{bmatrix}

Go to Question:

Prev Question

Next Question