If then

Engineering

Mathematics

Algebra of Matrices

Solving System of Linear Equation Cramers Rule

cubic and higher degree equation

Question

If $[\begin{array}{cc} 2 & - 3 \\ 1 & λ \end{array}] \times [\begin{array}{ccc} 1 & 5 & μ \\ 0 & 2 & - 3 \end{array}] = [\begin{array}{ccc} 2 & 4 & 1 \\ 1 & - 1 & 13 \end{array}]$ then

λ = 3, µ = – 4

λ = 4, µ = – 3

no real values of λ, µ are possible

none of these

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Given,

[\begin{array}{cc} 2 & - 3 \\ 1 & λ \end{array}] \times [\begin{array}{ccc} 1 & 5 & μ \\ 0 & 2 & - 3 \end{array}] = [\begin{array}{ccc} 2 & 4 & 1 \\ 1 & - 1 & 13 \end{array}]

(\begin{array}{ccc} 2 & 4 & 2 μ - 9 \\ 1 & 5 + 2 λ & μ - 3 λ \end{array}) = (\begin{array}{ccc} 2 & 4 & 1 \\ 1 & - 1 & 13 \end{array})

Comparing both sides we get

2µ + 9 = 1 ⇒ 2µ = – 8 ⇒ µ = – 4

Also 5 + 2λ = – 1 ⇒ 2λ = 6 ⇒ λ = 3

Hence, we get µ = – 4, λ = 3.