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Matrices and Determinants question

2016 · Shift 0 · Q29

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JEE Main
Mathematics
Matrices and Determinants
2016 · Shift 0 · Q29

Matrices and Determinants question

2016 · Shift 0 · Q29

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1

If

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

and

A

adj

A=A{A^T},

then

5a+b

is equal to :

A
$4$
B
$13$
C
$-1$
D
$5$

View written solutionFree

Correct answer: D

Interpret the condition

Given $A=\begin{bmatrix}5a & -b\\ 3 & 2\end{bmatrix}$ and $A\,\operatorname{adj}(A)=A A^T.$

For any square matrix $A$ , $A\,\operatorname{adj}(A)=|A|I.$ So the given condition becomes $|A|I = AA^T.$

Compute $|A|$

$|A|=\begin{vmatrix}5a & -b\\ 3 & 2\end{vmatrix}=10a+3b.$

Hence,

\begin{bmatrix}10a+3b&0\\0&10a+3b\end{bmatrix}.$$ --- 3. **Compute $AA^T$** First, $$A^T=\begin{bmatrix}5a&3\\-b&2\end{bmatrix}.$$ Now, $$AA^T= \begin{bmatrix}5a&-b\\3&2\end{bmatrix} \begin{bmatrix}5a&3\\-b&2\end{bmatrix}.

Multiplying,

\begin{bmatrix} 25a^2+b^2 & 15a-2b\\ 15a-2b & 13 \end{bmatrix}.$$ --- 4. **Equate corresponding entries** Since $$AA^T=|A|I,$$ we must have $$\begin{bmatrix} 25a^2+b^2 & 15a-2b\\ 15a-2b & 13 \end{bmatrix} = \begin{bmatrix}10a+3b&0\\0&10a+3b\end{bmatrix}.$$ So, - Off-diagonal entry: $$15a-2b=0 \implies 15a=2b \implies b=\frac{15a}{2}.$$ - Bottom-right entry: $$13=10a+3b.$$ Substitute $b=\frac{15a}{2}$: $$10a+3\left(\frac{15a}{2}\right)=13.$$ $$10a+\frac{45a}{2}=13.$$ $$\frac{20a+45a}{2}=13.$$ $$\frac{65a}{2}=13.$$ $$65a=26 \implies a=\frac{2}{5}.$$ Then $$b=\frac{15}{2}\cdot \frac{2}{5}=3.$$ --- 5. **Find $5a+b$** $$5a+b=5\left(\frac{2}{5}\right)+3=2+3=5.$$ --- 6. **Check with options** Thus, $$5a+b=5.$$ So the correct option is **D**. --- 7. **Compare with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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