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If A=[32a15]A = \begin{bmatrix} 3 & 2a \\ 1 & 5 \end{bmatrix} and B=[23b5]B = \begin{bmatrix} 2 & 3 \\ b & 5 \end{bmatrix} both are singular matrices, then a+ba + b is equal to

Solution

✅ Correct Option: 3

A singular matrix has determinant equal to zero.

For matrix A=[32a15]A = \begin{bmatrix} 3 & 2a \\ 1 & 5 \end{bmatrix}:

det(A) =(3)(5)−(2a)(1)= (3)(5) - (2a)(1)

det(A) =15−2a= 15 - 2a

Since AA is singular:

15−2a=015 - 2a = 0

2a=152a = 15

a=152a = \dfrac{15}{2}


For matrix B=[23b5]B = \begin{bmatrix} 2 & 3 \\ b & 5 \end{bmatrix}:

det(B) =(2)(5)−(3)(b)= (2)(5) - (3)(b)

det(B) =10−3b= 10 - 3b

Since BB is singular:

10−3b=010 - 3b = 0

3b=103b = 10

b=103b = \dfrac{10}{3}


a+b=152+103a + b = \dfrac{15}{2} + \dfrac{10}{3}

a+b=15×32×3+10×23×2a + b = \dfrac{15 \times 3}{2 \times 3} + \dfrac{10 \times 2}{3 \times 2}

a+b=456+206a + b = \dfrac{45}{6} + \dfrac{20}{6}

a+b=656a + b = \dfrac{65}{6}

Therefore, a+b=656a + b = \dfrac{65}{6}.

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