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MHTCET 25 Apr 2025 Shift 2 Paper

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Question : 106 of 150

Marks: +1, -0

If

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

A \cdot

A=\mathrm{AA}^{\mathrm{T}}

5 a+b=

7
9
13
5

Solution:

Concept:

For a square matrix

A

, the property

A \cdot \operatorname{adj} A = A A^T

helps compare entries after expanding both products.

Explanation:

For

A=\begin{bmatrix}5a & -b \\ 3 & 2\end{bmatrix}

, first find

\operatorname{adj} A

.

Cofactors are:

A_{11}=2

,

A_{12}=-3

,

A_{21}=b

,

A_{22}=5a

.

So

\operatorname{adj} A=\begin{bmatrix}2 & b \\ -3 & 5a\end{bmatrix}

.

Now compute both sides:

A \operatorname{adj} A=\begin{bmatrix}5a & -b \\ 3 & 2\end{bmatrix}\begin{bmatrix}2 & b \\ -3 & 5a\end{bmatrix}=\begin{bmatrix}10a+3b & 0 \\ 0 & 10a+3b\end{bmatrix}

Also,

A^T=\begin{bmatrix}5a & 3 \\ -b & 2\end{bmatrix}

, hence

A A^T=\begin{bmatrix}25a^2+b^2 & 15a-2b \\ 15a-2b & 13\end{bmatrix}

Equating corresponding entries gives:

15a-2b=0

and

10a+3b=13

.

From

15a-2b=0

, we get

b=\frac{15a}{2}

.

Substitute into

10a+3b=13

:

10a+3\left(\frac{15a}{2}\right)=13 \Rightarrow \frac{65a}{2}=13 \Rightarrow a=\frac{2}{5}

.

Then

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

.

Therefore,

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

.

Answer:

5a+b=5

, which is option D.

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