How to multiply a matrix by a scalar

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ACT Math › How to multiply a matrix by a scalar

Questions 1 - 8

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ , and let be the 3x3 identity matrix.

If , evaluate $n_{2,2}$ .

The correct answer is not given among the other responses.

Explanation

The 3x3 identity matrix is

$I = \begin{bmatrix} 1&0 &0 \ 0&1&0 \ 0&0 &1\end{bmatrix}$

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$m_{2,2} = n_{2,2} +3i_{2,2}$

$m_{2,2}$ is the second element in the second row, which is 6; similarly, $i_{2,2} = 1$ . The equation becomes

$6= n_{2,2} +3 (1)$

$6= n_{2,2} +3$

$n_{2,2} =3$

Simplify the following

$3\cdot \begin{bmatrix} -3& 5&0 \ 4& -1& 2 \end{bmatrix}$

$\begin{bmatrix} -9& 15&0 \ 12& -3& 6 \end{bmatrix}$

$\begin{bmatrix} -9& 5&0 \ 12& -1& 2 \end{bmatrix}$

$\begin{bmatrix} 6 \ 15 \end{bmatrix}$

$\begin{bmatrix} 0& 0&0 \ 1& 1& 1 \end{bmatrix}$

Explanation

When multplying any matrix by a scalar quantity (3 in our case), we simply multiply each term in the matrix by the scalar.

Therefore, every number simply gets multiplied by 3, giving us our answer.

$\begin{bmatrix} -9& 15&0 \ 12& -3& 6 \end{bmatrix}$

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ , and let be the 3x3 identity matrix.

If , then evaluate $n _{3,1}$ .

Explanation

The 3x3 identity matrix is

$I = \begin{bmatrix} 1&0 &0 \ 0&1&0 \ 0&0 &1\end{bmatrix}$

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$n _{3,1} = m _{3,1} -6 i _{3,1}$

$m _{3,1}$ is the first element in the third row of , which is 3; similarly, $i _{3,1} = 0$ . Therefore,

$n _{3,1} = 3 -6 (0)= 3$

$3*\begin{bmatrix} 0&3 \ -1&-2 \end{bmatrix}$

$\begin{bmatrix} 0&9 \ -3& -6 \end{bmatrix}$

$\begin{bmatrix} 0&9 \ -6&-3 \end{bmatrix}$

$\begin{bmatrix} 0&6 \ -3&-6 \end{bmatrix}$

$\begin{bmatrix} 0&-3 \ -9& -6 \end{bmatrix}$

$\begin{bmatrix} 0&-6 \ -6& -9 \end{bmatrix}$

Explanation

When multiplying a constant to a matrix, multiply each entry in the matrix by the constant.

$3*\begin{bmatrix} 0&3 \ -1&-2 \end{bmatrix}=$

$\begin{bmatrix} 0*3&3*3 \ -1*3& -2*3 \end{bmatrix}=$

$\begin{bmatrix} 0&9 \ -3& -6 \end{bmatrix}$

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ .

If , evaluate $n_{2,1}$ .

The correct answer is not among the other responses.

Explanation

If , then $N = \frac{1}{5} M$ .

Scalar multplication of a matrix is done elementwise, so

$n_{2,1} = \frac{1}{5} m_{2,1}$

$m_{2,1}$ is the first element in the second row of , which is 5, so

$n_{2,1} = \frac{1}{5} (5) = 1$

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ , and let be the 3x3 identity matrix.

If , evaluate $n_{2,1}$ .

$1\frac{1}{4}$

$-1\frac{1}{4}$

$-1\frac{1}{2}$

The correct answer is not given among the other responses.

Explanation

The 3x3 identity matrix is

$I = \begin{bmatrix} 1&0 &0 \ 0&1&0 \ 0&0 &1\end{bmatrix}$

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$m_{2,1} = 4n_{2,1} +i_{2,1}$

$m_{2,1}$ is the first element in the second row, which is 5; similarly, $i_{2,1} = 0$ . The equation becomes

$5= 4n_{2,1} +0$

$5= 4n_{2,1}$

$n_{2,1}= 5 \div 4 = 1\frac{1}{4}$

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ .

If , evaluate $n_{2,3}$ .

$\frac{1}{3}$

$-1\frac{1}{3}$

The correct answer is not among the other responses.

Explanation

Scalar multplication of a matrix is done elementwise, so

$n_{2,3} = 3 m _{2,3}$

$m _{2,3}$ is the third element in the second row of , which is 1, so

$n_{2,3} = 3 \cdot 1 = 3$

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ , and let be the 3x3 identity matrix.

If , then evaluate $n _{3,1}$ .

Explanation

The 3x3 identity matrix is

$I = \begin{bmatrix} 1&0 &0 \ 0&1&0 \ 0&0 &1\end{bmatrix}$

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$n _{3,1} = 7 m _{3,1} + i _{3,1}$

$m _{3,1}$ is the first element in the third row of , which is 3; similarly, $i _{3,1} = 0$ . Therefore,

$n _{3,1} = 7 (3) + 0 = 21$

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