# A=begin{bmatrix}2& 1&1 -1 & -1&4 end{bmatrix} B=begin{bmatrix}0& 2 -4 & 12 & -3 end{bmatrix} C=begin{bmatrix}6& -1 3 & 0-2 & 5 end{bmatrix} D=begin{bmatrix}2& -3&4 -3 & 1&-2 end{bmatrix} a)A-3D b)B+frac{1}{2} c) C+ frac{1}{2}B (a),(b),(c) need to be solved

$A=\left[\begin{array}{ccc}2& 1& 1\\ -1& -1& 4\end{array}\right]B=\left[\begin{array}{cc}0& 2\\ -4& 1\\ 2& -3\end{array}\right]C=\left[\begin{array}{cc}6& -1\\ 3& 0\\ -2& 5\end{array}\right]D=\left[\begin{array}{ccc}2& -3& 4\\ -3& 1& -2\end{array}\right]$
a)$A-3D$
b)$B+\frac{1}{2}$
c) $C+\frac{1}{2}B$
(a),(b),(c) need to be solved
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

un4t5o4v
Step 1
The addition or subtraction of two matrices A and B are possible only if the number of rows and columns of the matrices A and Bare equal.
Suppose that the matrices A and B have same number of rows and columns. Then, A+B is calculated by adding corresponding elements of the matrices A and B, and A-B is calculated by subtracting corresponding elements of the matrices B from A.
Step 2
a)We have, . So , $3D=\left[\begin{array}{ccc}3×2& 3×\left(-3\right)& 3×4\\ 3×\left(-3\right)& 3×1& 3×\left(-2\right)\end{array}\right]=\left[\begin{array}{ccc}6& -9& 12\\ -9& 3& -6\end{array}\right]$
Now ,
$A-3D=\left[\begin{array}{ccc}2& 1& 1\\ -1& -1& 4\end{array}\right]-\left[\begin{array}{ccc}6& -9& 12\\ -9& 3& -6\end{array}\right]$
$=\left[\begin{array}{ccc}2-6& 1-\left(-9\right)& 1-12\\ -1-\left(-9\right)& -1-3& 4-\left(-6\right)\end{array}\right]$
$=\left[\begin{array}{ccc}-4& 10& -11\\ 8& -4& 10\end{array}\right]$
Step 3
b. We have,
$B=\left[\begin{array}{cc}0& 2\\ -4& 1\\ 2& -3\end{array}\right]$ Since the matrix B contains 3 rows and 2 columns while $\frac{1}{2}$ is a scalar which is a matrix of one row and one column, so the operation $B+\frac{1}{2}$ is not possible. Because the addition or subtraction of two matrices A and B are possible only if the number of rows and columns of the matrices A and Bare equal.
Step 4
c) We have,
$B=\left[\begin{array}{cc}0& 2\\ -4& 1\\ 2& -3\end{array}\right]C=\left[\begin{array}{cc}6& -1\\ 3& 0\\ -2& 5\end{array}\right]$ So , $\frac{1}{2}B=\left[\begin{array}{cc}\frac{1}{2}×0& \frac{1}{2}×2\\ \frac{1}{2}×\left(-4\right)& \frac{1}{2}×1\\ \frac{1}{2}×2& \frac{1}{2}×\left(-3\right)\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -2& \frac{1}{2}\\ 1& -\frac{3}{2}\end{array}\right]$
Now,
$\left[\begin{array}{cc}6& -1\\ 3& 0\\ -2& 5\end{array}\right]+\left[\begin{array}{cc}0& 1\\ -2& \frac{1}{2}\\ 1& -\frac{3}{2}\end{array}\right]$
$=\left[\begin{array}{cc}6+0& -1+1\\ 3+\left(-2\right)& 0+\frac{1}{2}\\ -2+1& 5+\left(-\frac{3}{2}\right)\end{array}\right]$
$=\left[\begin{array}{cc}6& 0\\ 3-2& \frac{1}{2}\\ -1& 5-\frac{3}{2}\end{array}\right]$
$=\left[\begin{array}{cc}6& 0\\ 1& \frac{1}{2}\\ -1& \frac{2×5-3}{2}\end{array}\right]$
$=\left[\begin{array}{cc}6& 0\\ 1& \frac{1}{2}\\ -1& \frac{10-3}{2}\end{array}\right]$
$=\left[\begin{array}{cc}6& 0\\ 1& \frac{1}{2}\\ -1& \frac{7}{2}\end{array}\right]$
###### Not exactly what you’re looking for?
[10​12​]X=$\left[\begin{array}{ccc}3& 4& 2\\ 1& 0& 3\end{array}\right]$[31​40​23​]
###### Not exactly what you’re looking for?
Jeffrey Jordon

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee