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MHTCET 15 May 2024 Shift 2 Paper

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Question : 117 of 150

Marks: +1, -0

If

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

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

5 a+b

is equal to

-1
5
3
13

Solution:

Concept:

Use the matrix identity

A \cdot \operatorname{adj} A = |A| I

and equate it with

A A^{T}

.

Explanation:

For

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

, the determinant is

|A| = 10a + 3b

.

Since

A \cdot \operatorname{adj} A = |A| I

, the given condition reduces to

|A| I = A A^{T}

.

Compute

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

.

Equating this with

\begin{bmatrix} 10a+3b & 0 \\ 0 & 10a+3b \end{bmatrix}

gives the equations:

15a - 2b = 0

and

10a + 3b = 13

.

From

15a - 2b = 0

, we get

b = \frac{15a}{2}

.

Substituting 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 \cdot \frac{2}{5} + 3 = 5

.

Answer:

5a + b = 5

, i.e., option B.

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