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MHT CET 2021 Math Solved Paper

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Question : 67 of 100

Marks: +1, -0

If

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

A

A=AA^{T}

5a+b

5
4
13
$-1$

Solution:

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

and

A

adj

A=AA^{T}

\therefore\;\;

adj

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

Now,

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

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

and

A\cdot\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}

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

is given, so equating the two expression, we get

\begin{bmatrix} 25a^{2}+b^{2} & 15a-2b \\ 15a-2b & 13 \end{bmatrix}=\begin{bmatrix} 10a+3b & 0 \\ 0 & 10a+3b \end{bmatrix}

We have,

10a+3b=13

and

15a-2b=0

On solving, we get

a=\;\frac{2}{5}

and

b=3

\Rightarrow\;\; 5a+b=5\times\;\frac{2}{3}+3

\Rightarrow\;\; 5a+b=2+3

\Rightarrow\;\; 5a+b=5

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