If A=≤ft[{array}{ccc}3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7{array}] ; B=≤ft[{array}{ccc}-1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4{array}] then, (2 A+3 B)^{-1}= \_\_\_\_

Correct Answer is : [13000−14000126]≤ft[{array}{ccc}1/3 & 0 & 0 \\ 0 & -1/4 & 0 \\ 0 & 0 & 1/26{array}]31000−41000261

≤ft[{array}{ccc}1/3 & 0 & 0 \\ 0 & -1/4 & 0 \\ 0 & 0 & 1/26{array}]

≤ft[{array}{ccc}1/3 & 0 & 0 \\ 0 & 1/4 & 0 \\ 0 & 0 & 1/26{array}]

≤ft[{array}{ccc}1/3 & 0 & 0 \\ 0 & 1/4 & 0 \\ 0 & 0 & -1/26{array}]

≤ft[{array}{ccc}-1/3 & 0 & 0 \\ 0 & 1/4 & 0 \\ 0 & 0 & -1/26{array}]

Correct Answer is : [13000−14000126]≤ft[{array}{ccc}1/3 & 0 & 0 \\ 0 & -1/4 & 0 \\ 0 & 0 & 1/26{array}]31000−41000261

Test Index

MHTCET 17 Apr 2026 Shift 2 Paper

Question : 113 of 150

Marks: +1, -0

A=\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7\end{array}\right] ; B=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{array}\right]

(2 A+3 B)^{-1}=

\_\_\_\_

$\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{26}\end{array}\right]$
$\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{26}\end{array}\right]$
$\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & -\frac{1}{26}\end{array}\right]$
$\left[\begin{array}{ccc}-\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & -\frac{1}{26}\end{array}\right]$

Solution:

Concept:

For a diagonal matrix, the inverse is obtained by taking the reciprocal of each diagonal element.

Explanation:

Given:

A=\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{array}\right]

First, compute

2A

and

3B

2A=\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 14\end{array}\right]

3B=\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12\end{array}\right]

Adding them:

2A+3B=\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 26\end{array}\right]

Since this is a diagonal matrix, its inverse is:

(2A+3B)^{-1}=\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{26}\end{array}\right]

Answer:

The correct answer is Option A:

\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{26}\end{array}\right]

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