Single Choice

If $$A = \begin{bmatrix} 5a & -b\\ 3 & 2\end{bmatrix}$$ and $$A \ adj\ A = A A^{T}$$, then $$5a + b$$ is equal to

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

Solution

$$A. A^T = A. adjA$$ ..... $$(i)$$

$$A. adj A = |A|(I)$$ ........ $$(ii)$$ $$\left[A^{-1}=\frac{1}{|A|} adj(A)\right]$$

Now, $$det(A) = 10a+3b$$

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

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

$$det(A) |I|= \begin{bmatrix} 10a+3b & 0\\ 0 & 10a+3b\end{bmatrix}$$

From $$(i)$$ and $$(ii)$$

$$det (A) |I| = A.A^T$$

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

$$\implies 10a+3b=13=25a^2+b^2$$

And $$15a-2b=0$$

$$\implies 15a=2b\implies b=\dfrac{15a}{2}$$

Now, $$10a+3b=13$$

$$\implies 20a+45a=26$$ .... [By substituting the value of $$b$$]

$$\implies a=\dfrac {26}{65}$$

Now, $$b=\dfrac {15a}{2} = \dfrac {15 \times 26}{2 \times 65}=3$$

Thus, $$5a+b = 5 \left(\dfrac {26}{65}\right)+3=5$$

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