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  • Let A = [1 2 - 1 3] and a = 4, b = -2. Show that: (bA)^T = bA^T

Let A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right] and a = 4, b = -2.

Show that:
(bA)^T = bA^T    

Answers (1)

a) To prove: (bA)^T = bA^T

As, LHS = (bA)^T = (-2A)^T=(-2\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right])^T

(bA)^T = (-2A)^T=\left[\begin{array}{cc} -2 & -4 \\ 2 & -6 \end{array}\right]^T=\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]

 

\begin{aligned} &\text { Similarly, }\\ &\mathrm{RHS}=-2\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^{\mathrm{T}}=-2\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]\\ &\text { Hence proved }L H S=R H S=\left[\begin{array}{cc} -2 &2 \\ -4 & -6 \end{array}\right]\\ &\text { Then, }(b A)^{T}=b A^{\top} \ldots \text { proved } \end{aligned}

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