Recap C — Matrices

Question: Write down the order of $A=\left[\begin{array}{cc}4& -1\end{array}\right]$, $B=\left[\begin{array}{c}2\\ 7\\ 0\end{array}\right]$ and $C=\left[\begin{array}{ccc}1& 0& 3\\ 2& 5& 1\\ 0& 4& 6\end{array}\right]$.

Answer:

• A: $1$ row, $2$ columns → $1\times 2$
• B: $3$ rows, $1$ column → $3\times 1$
• C: $3$ rows, $3$ columns → $3\times 3$ (a square matrix)

Question: Find $A+B$ for $A=\left[\begin{array}{cc}3& 1\\ 4& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}5& 2\\ 1& 6\end{array}\right]$.

Answer:

Add matching positions: $\left[\begin{array}{cc}3+5& 1+2\\ 4+1& 2+6\end{array}\right]=\left[\begin{array}{cc}8& 3\\ 5& 8\end{array}\right]$

Question: Find $A-B$ for $A=\left[\begin{array}{ccc}6& 2& -1\\ 2& -4& 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}1& -3& 4\\ 5& 2& -2\end{array}\right]$.

Answer:

Subtract matching positions — take care with negatives: $6-1=5$, $2-(-3)=5$, $-1-4=-5$, $2-5=-3$, $-4-2=-6$, $0-(-2)=2$.

$A-B=\left[\begin{array}{ccc}5& 5& -5\\ -3& -6& 2\end{array}\right]$

Question: A is a $3\times 2$ matrix and B is a $3\times 3$ matrix. Find $A+B$.

Answer:

Not possible — A and B have different orders, so there are not matching positions for every number. Write exactly this as your answer.

Question: Find $3A$ for $A=\left[\begin{array}{cc}2& -5\\ 0& 4\end{array}\right]$.

Answer:

Multiply every number by $3$: $3A=\left[\begin{array}{cc}6& -15\\ 0& 12\end{array}\right]$

Question: Find $2A-B$ for $A=\left[\begin{array}{cc}3& 1\\ 2& 5\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 4\\ 6& 2\end{array}\right]$.

Answer:

• 1$2A=\left[\begin{array}{cc}6& 2\\ 4& 10\end{array}\right]$
• 2$2A-B=\left[\begin{array}{cc}5& -2\\ -2& 8\end{array}\right]$

Question: Find $AB$ for $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 0\\ 1& 3\end{array}\right]$.

Answer:

• Row $1$ × column $1$: $1\times 2+2\times 1=4$
• Row $1$ × column $2$: $1\times 0+2\times 3=6$
• Row $2$ × column $1$: $3\times 2+4\times 1=10$
• Row $2$ × column $2$: $3\times 0+4\times 3=12$

$AB=\left[\begin{array}{cc}4& 6\\ 10& 12\end{array}\right]$

Question: Using the same matrices, find $BA$. Is it the same as $AB$?

Answer:

• Row $1$ of B × column $1$ of A: $2\times 1+0\times 3=2$
• Row $1$ × column $2$: $2\times 2+0\times 4=4$
• Row $2$ × column $1$: $1\times 1+3\times 3=10$
• Row $2$ × column $2$: $1\times 2+3\times 4=14$

$BA=\left[\begin{array}{cc}2& 4\\ 10& 14\end{array}\right]$ — not the same as $AB$. The order of multiplication matters.

Question: Find $AB$ for $A=\left[\begin{array}{cc}2& 3\\ 1& -4\end{array}\right]$ and $B=\left[\begin{array}{c}5\\ 2\end{array}\right]$.

Answer:

Orders: ($2\times 2$) × ($2\times 1$) → answer is $2\times 1$.

• Row $1$: $2\times 5+3\times 2=16$
• Row $2$: $1\times 5+(-4)\times 2=-3$

$AB=\left[\begin{array}{c}16\\ -3\end{array}\right]$

Question: Find $AB$ for $A=\left[\begin{array}{ccc}1& 0& 2\\ 3& 1& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 1\\ 0& 3\\ 1& 4\end{array}\right]$.

Answer:

Orders: ($2\times 3$) × ($3\times 2$) — inside numbers match, the answer is $2\times 2$.

• Row $1$ × column $1$: $1(2)+0(0)+2(1)=4$
• Row $1$ × column $2$: $1(1)+0(3)+2(4)=9$
• Row $2$ × column $1$: $3(2)+1(0)+(-1)(1)=5$
• Row $2$ × column $2$: $3(1)+1(3)+(-1)(4)=2$

$AB=\left[\begin{array}{cc}4& 9\\ 5& 2\end{array}\right]$

Question: A is $2\times 3$ and B is $3\times 3$. Can you find $AB$? Can you find $BA$?

Answer:

• $AB$: ($2$ × $3$) × ($3$ × $3$) — inside numbers match → possible, order $2\times 3$.
• $BA$: ($3$ × $3$) × ($2$ × $3$) — $3$ ≠ $2$ → not possible.

Question: Find the determinant of $\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]$.

Answer:

$ad-bc=4\times 5-3\times 2=20-6=14$

Question: Find the determinant of $\left[\begin{array}{cc}-3& 2\\ 4& -1\end{array}\right]$.

Answer:

$(-3)(-1)-(2)(4)=3-8=-5$

Question: Find the determinant of $\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$.

Answer:

$2\times 6-4\times 3=12-12=0$. A determinant of $0$ means the matrix is singular (it has no inverse).

Question: Find the determinant of $\left[\begin{array}{ccc}2& 1& 3\\ 0& 4& 1\\ 1& -2& 2\end{array}\right]$.

Answer:

Top row: $2$, $1$, $3$ with signs + − +.

$=2(4\times 2-1\times (-2))-1(0\times 2-1\times 1)+3(0\times (-2)-4\times 1)$

$=2(10)-1(-1)+3(-4)=20+1-12=9$

Question: Find the determinant of $\left[\begin{array}{ccc}1& -2& 3\\ 2& 0& -1\\ 4& 1& 2\end{array}\right]$.

Answer:

Top row: $1$, $-2$, $3$. The middle sign is −, and the number is $-2$, so that term becomes $-(-2)=+2$.

$=1(0\times 2-(-1)\times 1)-(-2)(2\times 2-(-1)\times 4)+3(2\times 1-0\times 4)$

$=1(1)-(-2)(8)+3(2)=1+16+6=23$

Question: Find $x$ if $\det \left[\begin{array}{cc}x& 3\\ 2& 5\end{array}\right]=4$.

Answer:

$5x-6=4\Rightarrow 5x=10\Rightarrow x=2$

$1$. $\left[\begin{array}{cc}4& -4\\ 4& 5\end{array}\right]$

$2$. $\left[\begin{array}{cc}4& 9\\ 6& 3\end{array}\right]$

$3$. $6\times 4-2\times 3=18$