Matrix Operations with Transpose

Linear algebra problems and solutions

Problem 636

Calculate the following expressions, using the following matrices:
\[A = \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix}, \qquad \mathbf{v} = \begin{bmatrix} 2 \\ -4 \end{bmatrix}\]

(a) $A B^\trans + \mathbf{v} \mathbf{v}^\trans$.

(b) $A \mathbf{v} – 2 \mathbf{v}$.

(c) $\mathbf{v}^{\trans} B$.

(d) $\mathbf{v}^\trans \mathbf{v} + \mathbf{v}^\trans B A^\trans \mathbf{v}$.

 
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Solution.

(a) $A B^\trans + \mathbf{v} \mathbf{v}^\trans$

\begin{align*} A B^\trans + \mathbf{v} \mathbf{v}^\trans &= \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & -1 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \end{bmatrix} \begin{bmatrix} 2 & -4 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -3 & -1 \\ -1 & -6 \end{bmatrix} + \begin{bmatrix} 4 & -8 \\ -8 & 16 \end{bmatrix} \\[6pt] &= \begin{bmatrix} 1 & -9 \\ -9 & 10 \end{bmatrix} \end{align*}

(b) $A \mathbf{v} – 2 \mathbf{v} $

\begin{align*} A \mathbf{v} – 2 \mathbf{v} &= \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \end{bmatrix} – 2 \begin{bmatrix} 2 \\ -4 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -8 \\ -14 \end{bmatrix} + \begin{bmatrix} -4 \\ 8 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -12 \\ -6 \end{bmatrix} \end{align*}

(c) $\mathbf{v}^{\trans} B$

\begin{align*} \mathbf{v}^{\trans} B &= \begin{bmatrix} 2 & -4 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -4 & 2 \end{bmatrix} \end{align*}

(d) $\mathbf{v}^\trans \mathbf{v} + \mathbf{v}^\trans B A^\trans \mathbf{v}$

\begin{align*}
&\mathbf{v}^\trans \mathbf{v} + \mathbf{v}^\trans B A^\trans \mathbf{v}\\[6pt] &= \begin{bmatrix} 2 & -4 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \end{bmatrix} + \left( \begin{bmatrix} 2 & -4 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \right) \left( \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \end{bmatrix} \right) \\[6pt] &= 20 + \begin{bmatrix} -4 & 2 \end{bmatrix} \begin{bmatrix} 24 \\ 2 \end{bmatrix} \\[6pt] &= 20 – 92 \\
&= -72 \end{align*}


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2 Responses

  1. Pavel says:

    Hello.

    I think that in the last expression $v^Tv$ must be 2*2+(-4)(-4)=4+16=20, not 18. And therefore answer is -72

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