Test Index

MHTCET 16 Apr 2026 Shift 1 Paper

© examsnet.com

Question : 115 of 150

Marks: +1, -0

If

A=\left[\begin{array}{cc}5 a & -b \\ 3 & 2\end{array}\right]

A(\operatorname{adj} A)=A A^{T}

5 a+b=

2
3
5
$\frac{15}{2}$

Solution:

Concept:

For any square matrix,

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

.

Compare this with the given equation

A(\operatorname{adj} A) = AA^T

.

Explanation:

Given

A=\left[\begin{array}{cc}5a & -b \\ 3 & 2\end{array}\right]

.

The determinant is

|A| = (5a)(2) - (-b)(3) = 10a + 3b

.

Therefore,

A(\operatorname{adj} A) = |A|I = \left[\begin{array}{cc}10a+3b & 0 \\ 0 & 10a+3b\end{array}\right]

.

Now compute

A^T = \left[\begin{array}{cc}5a & 3 \\ -b & 2\end{array}\right]

.

So,

AA^T = \left[\begin{array}{cc}5a & -b \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}5a & 3 \\ -b & 2\end{array}\right] = \left[\begin{array}{cc}25a^2+b^2 & 15a-2b \\ 15a-2b & 13\end{array}\right]

.

Equating

A(\operatorname{adj} A)

and

AA^T

gives the off-diagonal equation

15a - 2b = 0

.

Hence,

b = \frac{15a}{2}

.

Equating the bottom-right entries gives

10a + 3b = 13

.

Substitute

b = \frac{15a}{2}

:

10a + \frac{45a}{2} = 13

.

This simplifies to

\frac{65a}{2} = 13

, so

a = \frac{2}{5}

.

Then

b = \frac{15}{2} \times \frac{2}{5} = 3

.

Finally,

5a + b = 5\left(\frac{2}{5}\right) + 3 = 5

.

Answer:

\boxed{5}

, i.e. option C.

© examsnet.com

Go to Question:

Prev Question

Next Question