Calculate expressions of the form c.A+B - High school Algebra I Exercises

Carry out the following operations.

3\times\begin{pmatrix} 3 & -2 & -4 \cr\cr 1 & 0 & -4 \end{pmatrix}-\begin{pmatrix} 0 & 1 & 4 \cr\cr 2 & -6 & -3 \end{pmatrix}

\begin{pmatrix} 9 & -7 & 4 \cr\cr 1 & 8 & 9 \end{pmatrix}

\begin{pmatrix} 8 & -7 & -4 \cr\cr 12 & 4 & -9 \end{pmatrix}

\begin{pmatrix} 2 & -4 & 6 \cr\cr -1 & 2 & 9 \end{pmatrix}

\begin{pmatrix} 9 & -7 & -16 \cr\cr 1 & 6 & -9 \end{pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

3 A=3 \cdot \begin {pmatrix} 3 & -2 & -4 \cr \cr 1 & 0 & 4 \end {pmatrix}=\begin {pmatrix} 9 & -6 & -12 \cr \cr \ 3 & 0 & -12 \end {pmatrix}

And:

\left(3 \cdot A\right)-B =\begin {pmatrix} 9 & -6 & -12 \cr \cr \ 3 & 0 & -12 \end {pmatrix}- \begin {pmatrix} 0 & 1 & 4 \cr \cr 2 & -6 & -3 \end {pmatrix}= \begin {pmatrix} 9 & -7 & -16 \cr \cr 1 & 6 & -9 \end {pmatrix}

3\times\begin{pmatrix} 3 & -2 & -4 \cr\cr 1 & 0 & -4 \end{pmatrix}-\begin{pmatrix} 0 & 1 & 4 \cr\cr 2 & -6 & -3 \end{pmatrix}=\begin{pmatrix} 9 & -7 & -16 \cr\cr 1 & 6 & -9 \end{pmatrix}

2\times\begin{pmatrix} 1 & 1 & 2 \cr\cr 2 & 0 & -1 \end{pmatrix}-\begin{pmatrix} 1 & 0 & 1 \cr\cr 1 & -6 & -2 \end{pmatrix}

\begin {pmatrix} 12 & 2 & 11 \cr \cr 1 & 2 & 10 \end {pmatrix}

\begin {pmatrix} 12 & -4 & -1 \cr \cr 3 & -5 & 0 \end {pmatrix}

\begin {pmatrix} 1 & 2 & 3 \cr \cr 3 & 6 & 0 \end {pmatrix}

\begin {pmatrix} 2 & 4 & 1 \cr \cr 13 & 5 & 0 \end {pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

2 A=2 \cdot \times\begin{pmatrix} 1 & 1 & 2 \cr\cr 2 & 0 & -1 \end{pmatrix}=\begin {pmatrix} 2 & 2 & 4 \cr \cr \ 4 & 0 & -2 \end {pmatrix}

And:

\left(2 \cdot A\right)-B =\begin {pmatrix} 2 & 2 & 4 \cr \cr \ 4 & 0 & -2 \end {pmatrix} - \begin{pmatrix} 1 & 0 & 1 \cr\cr 1 & -6 & -2 \end{pmatrix}= \begin {pmatrix} 1 & 2 & 3 \cr \cr 3 & 6 & 0 \end {pmatrix}

2\times\begin{pmatrix} 1 & 1 & 2 \cr\cr 2 & 0 & -1 \end{pmatrix}-\begin{pmatrix} 1 & 0 & 1 \cr\cr 1 & -6 & -2 \end{pmatrix}=\begin {pmatrix} 1 & 2 & 3 \cr \cr 3 & 6 & 0 \end {pmatrix}

2\times\begin{pmatrix} 4 & 3 & 3 \cr\cr 2 & 5 & 3 \end{pmatrix}-\begin{pmatrix} 1 & 2 & 3 \cr\cr 4 & 0 & 0 \end{pmatrix}

\begin {pmatrix} 7& 4 & 3 \cr \cr 0 & 10 & 6 \end {pmatrix}

\begin {pmatrix} 3& 5 & 1\cr \cr 10 & 4 & 5 \end {pmatrix}

\begin {pmatrix} 3& 12 & 9 \cr \cr 4 & 0 & 5 \end {pmatrix}

\begin {pmatrix} 5& 2 & 6 \cr \cr 8 & 1 & 3 \end {pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

2\times\begin{pmatrix} 4 & 3 & 3 \cr\cr 2 & 5 & 3 \end{pmatrix}=\begin{pmatrix} 8 & 6 & 6 \cr\cr 4 & 10 & 6 \end{pmatrix}

And:

\left(2 \cdot A\right)-B =\begin{pmatrix} 8 & 6 & 6 \cr\cr 4 & 10 & 6 \end{pmatrix} - \begin{pmatrix} 1 & 2 & 3 \cr\cr 4 & 0 & 0 \end{pmatrix}= \begin {pmatrix} 7& 4 & 3 \cr \cr 0 & 10 & 6 \end {pmatrix}

\left(2 \cdot A\right)-B = \begin {pmatrix} 7& 4 & 3 \cr \cr 0 & 10 & 6 \end {pmatrix}

3\times\begin{pmatrix} 1 & 1 & 0 \cr\cr 0 & 0 & 4 \end{pmatrix}-\begin{pmatrix} 0 & 1 & 2 \cr\cr 1 & -2 & 1 \end{pmatrix}

\begin {pmatrix} 12& 1 & 8 \cr \cr 2 & 2 &10 \end {pmatrix}

\begin {pmatrix} 3& 2 & -2 \cr \cr -1 & 2 &11 \end {pmatrix}

\begin {pmatrix} 2& -2 & 2 \cr \cr -10 & 3 &9 \end {pmatrix}

\begin {pmatrix} 7& 1 & 2 \cr \cr -5& -2 &1 \end {pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

3\times\begin{pmatrix} 1 & 1 & 0 \cr\cr 0 & 0 & 4 \end{pmatrix}=\begin{pmatrix} 3 & 3 & 0 \cr\cr 0 & 0 & 12 \end{pmatrix}

And:

\left(3 \cdot A\right)-B =\begin{pmatrix} 3 & 3 & 0 \cr\cr 0 & 0 & 12 \end{pmatrix} - \begin{pmatrix} 0 & 1 & 2 \cr\cr 1 & -2 & 1 \end{pmatrix}= \begin {pmatrix} 3& 2 & -2 \cr \cr -1 & 2 &11 \end {pmatrix}

\left(3 \cdot A\right)-B = \begin {pmatrix} 3& 2 & -2 \cr \cr -1 & 2 &11 \end {pmatrix}

2\times\begin{pmatrix} 1 & -2 \cr\cr 2 & 0 \end{pmatrix}-\begin{pmatrix} 0 & 1 \cr\cr 1 & 0 \end{pmatrix}

\begin {pmatrix} 7& 3 \cr \cr 2 & 6 \end {pmatrix}

\begin {pmatrix} 2& -5 \cr \cr 3 & 0 \end {pmatrix}

\begin {pmatrix} 4& 3 \cr \cr 1 & 3 \end {pmatrix}

\begin {pmatrix} 7& 2 \cr \cr 9 & -3 \end {pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

2\times\begin{pmatrix} 1 & -2 \cr\cr 2 & 0 \end{pmatrix}=\begin{pmatrix} 2 & -4 \cr\cr 4 & 0 \end{pmatrix}

And:

\left(2 \cdot A\right)-B =\begin{pmatrix} 2 & -4 \cr\cr 4 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 \cr\cr 1 & 0 \end{pmatrix}= \begin {pmatrix} 2& -5 \cr \cr 3 & 0 \end {pmatrix}

2\times\begin{pmatrix} 1 & -2 \cr\cr 2 & 0 \end{pmatrix}-\begin{pmatrix} 0 & 1 \cr\cr 1 & 0 \end{pmatrix}=\begin {pmatrix} 2& -5 \cr \cr 3 & 0 \end {pmatrix}

2\times\begin{pmatrix} 1 & 1 \cr\cr 1 & 0 \end{pmatrix}-\begin{pmatrix} 1 & 2 \cr\cr 1 & 0 \end{pmatrix}

\begin {pmatrix} 5& 4 \cr \cr 2 & 1 \end {pmatrix}

\begin {pmatrix} 1& 0 \cr \cr 1 & 0 \end {pmatrix}

\begin {pmatrix} 3& 2 \cr \cr 2 & -1 \end {pmatrix}

\begin {pmatrix} -1& 2 \cr \cr 3 & -2 \end {pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

2\times\begin{pmatrix} 1 & 1 \cr\cr 1 & 0 \end{pmatrix}=\begin{pmatrix} 2 & 2 \cr\cr 2 & 0 \end{pmatrix}

And:

\left(2 \cdot A\right)-B =\begin{pmatrix} 2 & 2 \cr\cr 2 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 2 \cr\cr 1 & 0 \end{pmatrix}= \begin {pmatrix} 1& 0 \cr \cr 1 & 0 \end {pmatrix}

2\times\begin{pmatrix} 1 & 1 \cr\cr 1 & 0 \end{pmatrix}-\begin{pmatrix} 1 & 2 \cr\cr 1 & 0 \end{pmatrix}=\begin {pmatrix} 1& 0 \cr \cr 1 & 0 \end {pmatrix}

3\times\begin{pmatrix} 1 & -1 \cr\cr -1 & 1 \end{pmatrix}-\begin{pmatrix} 1 & -1 \cr\cr -1 & 0 \end{pmatrix}

\begin {pmatrix} 3& -1 \cr \cr 0 & 1 \end {pmatrix}

\begin {pmatrix} 1& -2 \cr \cr 0 & -1 \end {pmatrix}

\begin {pmatrix} 2& -2 \cr \cr -2 & 0 \end {pmatrix}

\begin {pmatrix} 1& -1 \cr \cr 2 & 1 \end {pmatrix}

When :

A = \left[{\begin {array} {c} a_ {i, j} \end {array}} \right]
B = \left[{\begin {array} {c} b_ {i, j} \end {array}} \right]
c is a real number

Then:

cA=\left[{\begin {array} {c} c \cdot a_ {i, j} \end {array}} \right]
A-B=\left[{\begin {array} {c} a_ {i, j} -b_ {i, j} \end {array}} \right]

Therefore we have :

3\times\begin{pmatrix} 1 & -1 \cr\cr -1 & 1 \end{pmatrix}=\begin{pmatrix} 3 & -3 \cr\cr -3 & 0 \end{pmatrix}

And:

\left(3\cdot A\right)-B =\begin{pmatrix} 3 & -3 \cr\cr -3 & 0 \end{pmatrix} - \begin{pmatrix} 1 & -1 \cr\cr -1 & 0 \end{pmatrix}= \begin {pmatrix} 2& -2 \cr \cr -2 & 0 \end {pmatrix}

3\times\begin{pmatrix} 1 & -1 \cr\cr -1 & 1 \end{pmatrix}-\begin{pmatrix} 1 & -1 \cr\cr -1 & 0 \end{pmatrix}=\begin {pmatrix} 2& -2 \cr \cr -2 & 0 \end {pmatrix}