Multiplying Matrices 1

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Question 2 of 4

Solve for `-1/2A` given that:

`A=[[6,-14],[18,10]]`

Question 3 of 4

Solve

`[[1,3,2]]times[[6,1,0],[0,3,1],[1,2,2]]`

Question 4 of 4

Solve

`[[2,6,-2,7],[1,0,3,1]]times[[3],[-1],[4],[6]]`

`-1/2A`	`=`	`-1/2[[6,-14],[18,10]]`	Substitute `A`

	`=`	\begin{bmatrix} -\frac{1}{2}\times6 & -\frac{1}{2}\times-14 \\[0.3em] -\frac{1}{2}\times18 & -\frac{1}{2}\times10 \\ \end{bmatrix}	Distribute `-1/2`

	`=`	`[[-3,7],[-9,-5]]`

		`[[1,3,2]]times``[[6,1,0],[0,3,1],[1,2,2]]`

	`=`	\begin{bmatrix} ((1\cdot\color{#9a00c7}{6})+(3\cdot\color{#9a00c7}{0})+(2\cdot\color{#9a00c7}{1})) & ((1\cdot\color{#9a00c7}{1})+(3\cdot\color{#9a00c7}{3})+(2\cdot\color{#9a00c7}{2})) & ((1\cdot\color{#9a00c7}{0})+(3\cdot\color{#9a00c7}{1})+(2\cdot\color{#9a00c7}{2})) \\ \end{bmatrix}

	`=`	`[[6+0+2,1+9+4,0+3+4]]`

	`=`	`[[8,14,7]]`

		`[[2,6,-2,7],[1,0,3,1]]times``[[3],[-1],[4],[6]]`

	`=`	\begin{bmatrix} (2\cdot\color{#9a00c7}{3})+(6\cdot\color{#9a00c7}{-1})+(-2\cdot\color{#9a00c7}{4})+(7\cdot\color{#9a00c7}{6}) \\[0.3em] (1\cdot\color{#9a00c7}{3})+(0\cdot\color{#9a00c7}{-1})+(3\cdot\color{#9a00c7}{4})+(1\cdot\color{#9a00c7}{6}) \end{bmatrix}

	`=`	`[[6+(-6)+(-8)+42],[3+0+12+6]]`

	`=`	`[[34],[21]]`