Multiply Matrices by Scalar: CCSS.Math.Content.HSN-VM.C.7 - Common Core: High School - Number and Quantity

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Question

Compute

$3\begin{bmatrix} 2 & 4&-2 \ 1&20 &-4 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$3\begin{bmatrix} 2 & 4&-2 \ 1&20 &-4 \end{bmatrix}=\begin{bmatrix} 2\cdot3 & 4\cdot3&-2\cdot3 \ 1\cdot3&20\cdot3 &-4\cdot3 \end{bmatrix}=\begin{bmatrix} 6 & 12&-6 \ 3&60 &-12 \end{bmatrix}$

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Question

Compute

$-17\begin{bmatrix} -9 & -12&16\ 0&-7 &10 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$-17\begin{bmatrix} -9 & -12&16\ 0&-7 &10 \end{bmatrix}=\begin{bmatrix} -17\cdot(-9) & -17\cdot(-12)&-17\cdot16 \ -17\cdot0&-17\cdot-7 &-17\cdot10 \end{bmatrix}=\begin{bmatrix} 153 & 204&-272 \ 0&119 &-170 \end{bmatrix}$

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Question

Compute

$-19\begin{bmatrix} 4 & 0&17 \ 7&15 &-9 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$-19\begin{bmatrix} 4 & 0&17 \ 7&15 &-9 \end{bmatrix}=\begin{bmatrix} -19\cdot4 & -19\cdot0&-19\cdot17 \ -19\cdot7&-19\cdot15 &-19\cdot(-9) \end{bmatrix}=\begin{bmatrix} -76& 0&-323 \ -133&-285 &171 \end{bmatrix}$

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Question

Compute

$21\begin{bmatrix} 9 & 3-&2 \ 14&0 &14 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$21\begin{bmatrix} 9 & -3&2 \ 14&0 &14 \end{bmatrix}=\begin{bmatrix} 21\cdot9 & 21\cdot(-3)&21\cdot2 \ 21\cdot14&21\cdot0 &21\cdot14 \end{bmatrix}=\begin{bmatrix} 189 & -63&42 \ 294&0 &294 \end{bmatrix}$

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Question

Compute

$5\begin{bmatrix} 20 & 4&-3 \ 2&14 &-1 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$5\begin{bmatrix} 20 & 4&-3 \ 2&14 &-1 \end{bmatrix}=\begin{bmatrix} 5\cdot20 & 5\cdot4&5\cdot(-3) \ 5\cdot2&5\cdot14 &5\cdot(-1) \end{bmatrix}=\begin{bmatrix} 100 & 20&-15 \ 10&70 &-5 \end{bmatrix}$

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Question

Compute

$7\begin{bmatrix} 7 & 12&-3 \ 17&3 &6 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$7\begin{bmatrix} 7 & 12&-3 \ 17&3 &6 \end{bmatrix}=\begin{bmatrix} 7\cdot7 & 7\cdot12&7\cdot(-3) \ 7\cdot17&7\cdot3 &7\cdot6 \end{bmatrix}=\begin{bmatrix} 49 & 84&-21 \ 119&21 &42 \end{bmatrix}$

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Question

Compute

$22\begin{bmatrix} 7 & 6&16 \ 0&0 &12 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$22\begin{bmatrix} 7 & 6&16 \ 0&0 &12 \end{bmatrix}=\begin{bmatrix} 22\cdot7 & 22\cdot6&22\cdot16 \ 22\cdot0&22\cdot0 &22\cdot12 \end{bmatrix}=\begin{bmatrix} 154 & 132&352 \ 0&0 &264 \end{bmatrix}$

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Question

Compute

$-4\begin{bmatrix} 5 & 2&12 \ 3&-1 &-2 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$-4\begin{bmatrix} 5 & 2&12 \ 3&-1 &-2 \end{bmatrix}=\begin{bmatrix} -4\cdot5 & -4\cdot2&-4\cdot12 \ -4\cdot3&-4\cdot(-1) &-4\cdot(-2) \end{bmatrix}=\begin{bmatrix} -20 & -8&-48 \ -12&4 &8 \end{bmatrix}$

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Question

Compute

$4\begin{bmatrix} 10 & -4&5 \ 6&-1 &1 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$4\begin{bmatrix} 10 & -4&5 \ 6&-1 &1 \end{bmatrix}=\begin{bmatrix} 4\cdot10 & 4\cdot(-4)&4\cdot5 \ 4\cdot6&4\cdot(-1) &4\cdot1 \end{bmatrix}=\begin{bmatrix} 40 & -16&20 \ 24&-4 &4 \end{bmatrix}$

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Question

Compute

$14\begin{bmatrix} 17 & -2&3 \ -5&-1 &2 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$14\begin{bmatrix} 17 & -2&3 \ -5&-1 &2 \end{bmatrix}=\begin{bmatrix} 14\cdot17 & 14\cdot(-2)&14\cdot3 \ 14\cdot(-5)&14\cdot(-1) &14\cdot2 \end{bmatrix}=\begin{bmatrix} 238 & -28&42 \ -70&-14 &28 \end{bmatrix}$

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Question

Compute

$22\begin{bmatrix} 1 & 1&0 \ 14&1 &10 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$22\begin{bmatrix} 1 & 1&0 \ 14&1 &10 \end{bmatrix}=\begin{bmatrix} 22\cdot1 & 22\cdot1&22\cdot0 \ 22\cdot14&22\cdot1 &22\cdot10 \end{bmatrix}=\begin{bmatrix} 22 & 22&0 \ 308&22 &220 \end{bmatrix}$

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Question

Compute

$25\begin{bmatrix} -4 & 18&11 \ 8&0 &1 \end{bmatrix}$

Answer

In order to compute this, we need to multiply each entry by the scalar .

$25\begin{bmatrix} -4 & 18&11 \ 8&0 &1 \end{bmatrix}=\begin{bmatrix} 25\cdot(-4) & 25\cdot18&25\cdot11 \ 25\cdot8&25\cdot0 &-25\cdot1 \end{bmatrix}=\begin{bmatrix} -100 & 450&275\ 200&0 &25 \end{bmatrix}$

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