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Mathematics

Given, A = [3153]\begin{bmatrix} 3 & 1 \ 5 & 3 \end{bmatrix} and B = [−1a3−5]\begin{bmatrix} -1 & a \ 3 & -5 \end{bmatrix} and product AB = [b745]\begin{bmatrix} b & 7 \ 4 & 5 \end{bmatrix}. Find the values of ‘a’ and ‘b’.

Matrices

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Answer

Given,

A = [3153]\begin{bmatrix} 3 & 1 \ 5 & 3 \end{bmatrix}

B = [−1a3−5]\begin{bmatrix} -1 & a \ 3 & -5 \end{bmatrix}

AB = [b745]\begin{bmatrix} b & 7 \ 4 & 5 \end{bmatrix}

Solving,

⇒AB=[3153]×[−1a3−5]⇒AB=[3(−1)+1(3)3(a)+1(−5)5(−1)+3(3)5(a)+3(−5)]⇒AB=[−3+33a−5−5+95a−15]⇒[b745]=[03a−545a−15]\Rightarrow AB = \begin{bmatrix} 3 & 1 \ 5 & 3 \end{bmatrix} \times \begin{bmatrix} -1 & a \ 3 & -5 \end{bmatrix} \\[1em] \Rightarrow AB = \begin{bmatrix} 3(-1) + 1(3) & 3(a) + 1(-5) \ 5(-1) + 3(3) & 5(a) + 3(-5) \end{bmatrix} \\[1em] \Rightarrow AB = \begin{bmatrix} -3 + 3 & 3a - 5 \ -5 + 9 & 5a - 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} b & 7 \ 4 & 5 \end{bmatrix} = \begin{bmatrix} 0 & 3a - 5 \ 4 & 5a - 15 \end{bmatrix} \\[1em]

Comparing corresponding elements,

∴ b = 0

⇒ 3a - 5 = 7

⇒ 3a = 7 + 5

⇒ 3a = 12

⇒ a = 123\dfrac{12}{3}

⇒ a = 4

Hence, a = 4 and b = 0.

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