Matrix Algebra: Addition, Scalar Multiplication & Transpose

The first operations of matrix arithmetic — and the algebraic laws that make matrices behave like numbers

For $A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & 5 & 6 \end{bmatrix}$: $a_{11}=1$, $a_{12}=2$, $a_{13}=-1$, $a_{21}=0$, $a_{22}=5$, $a_{23}=6$.

$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} + \begin{bmatrix} 4 & 2 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1+4 & -1+2 \\ 2+(-1) & 3+0 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 1 & 3 \end{bmatrix}.$$

$$\begin{bmatrix}1&0&-2\\3&1&4\end{bmatrix} +\begin{bmatrix}0&2&1\\-1&0&2\end{bmatrix} +\begin{bmatrix}2&-1&0\\0&3&-3\end{bmatrix} =\begin{bmatrix}3&1&-1\\2&4&3\end{bmatrix}.$$

Find $a$ and $b$ if

$$\begin{bmatrix}a & 2\\1 & b\end{bmatrix} +\begin{bmatrix}3 & -1\\0 & 4\end{bmatrix} =\begin{bmatrix}5 & 1\\1 & 6\end{bmatrix}.$$

$\textit{Solution.}$ Entry-by-entry: $(1,1)$: $a+3=5 \Rightarrow a=2$; $\quad$ $(2,2)$: $b+4=6 \Rightarrow b=2$.

Find matrix $X$ if $X + \begin{bmatrix}2&1\\-1&3\end{bmatrix} = \begin{bmatrix}5&0\\2&1\end{bmatrix}.$

$\textit{Solution.}$ Subtract the known matrix from both sides (entry by entry):

$$X = \begin{bmatrix}5&0\\2&1\end{bmatrix} - \begin{bmatrix}2&1\\-1&3\end{bmatrix} = \begin{bmatrix}3&-1\\3&-2\end{bmatrix}.$$

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

Let $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$ and $B=\begin{bmatrix}5&6\\7&8\end{bmatrix}$.

$$A+B=\begin{bmatrix}6&8\\10&12\end{bmatrix}=B+A. \quad\checkmark$$

$$3\begin{bmatrix}1&-2\\0&4\end{bmatrix} = \begin{bmatrix}3\cdot1&3\cdot(-2)\\3\cdot0&3\cdot4\end{bmatrix} = \begin{bmatrix}3&-6\\0&12\end{bmatrix}.$$

$$-2\begin{bmatrix}1&3\\-1&5\end{bmatrix} =\begin{bmatrix}-2&-6\\2&-10\end{bmatrix}.$$

$$\frac{1}{2}\begin{bmatrix}4&-6\\2&8\end{bmatrix} =\begin{bmatrix}2&-3\\1&4\end{bmatrix}.$$

Let $A=\begin{bmatrix}1&0\\-1&2\end{bmatrix}$ and $B=\begin{bmatrix}2&-1\\3&0\end{bmatrix}$. Find $2A-3B$.

$$2A = \begin{bmatrix}2&0\\-2&4\end{bmatrix}, \qquad 3B = \begin{bmatrix}6&-3\\9&0\end{bmatrix}.$$

$$2A-3B = \begin{bmatrix}2-6&0-(-3)\\-2-9&4-0\end{bmatrix} = \begin{bmatrix}-4&3\\-11&4\end{bmatrix}.$$

Let $k=3$, $A=\begin{bmatrix}1&2\\0&-1\end{bmatrix}$, $B=\begin{bmatrix}-1&0\\2&3\end{bmatrix}$.

$\textit{Method 1 (compute $A+B$ first, then multiply):}$

$$A+B=\begin{bmatrix}0&2\\2&2\end{bmatrix}, \qquad 3(A+B)=\begin{bmatrix}0&6\\6&6\end{bmatrix}.$$

$\textit{Method 2 (distribute first):}$

$$3A=\begin{bmatrix}3&6\\0&-3\end{bmatrix},\quad 3B=\begin{bmatrix}-3&0\\6&9\end{bmatrix},\quad 3A+3B=\begin{bmatrix}0&6\\6&6\end{bmatrix}. \checkmark$$

Show that $(2+5)A = 2A+5A$ for $A=\begin{bmatrix}1&-1\\3&0\end{bmatrix}$.

$$7A=\begin{bmatrix}7&-7\\21&0\end{bmatrix}, \qquad 2A+5A=\begin{bmatrix}2&-2\\6&0\end{bmatrix} +\begin{bmatrix}5&-5\\15&0\end{bmatrix} =\begin{bmatrix}7&-7\\21&0\end{bmatrix}. \checkmark$$

Show that $(2\cdot3)A = 2(3A)$ for $A=\begin{bmatrix}1&2\\-1&4\end{bmatrix}$.

$$6A=\begin{bmatrix}6&12\\-6&24\end{bmatrix}, \qquad 3A=\begin{bmatrix}3&6\\-3&12\end{bmatrix},\quad 2(3A)=\begin{bmatrix}6&12\\-6&24\end{bmatrix}. \checkmark$$

$$A = \begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}_{2\times3} \implies A^{\mathsf{T}} = \begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}_{3\times2}.$$

Row 1 of $A$ becomes column 1 of $A^{\mathsf{T}}$; row 2 becomes column 2.

$$\begin{bmatrix}1&-1\\2&3\end{bmatrix}^{\mathsf{T}} =\begin{bmatrix}1&2\\-1&3\end{bmatrix}, \qquad \begin{bmatrix}1\\2\\3\end{bmatrix}^{\mathsf{T}} =\begin{bmatrix}1&2&3\end{bmatrix}, \qquad \begin{bmatrix}1&2&-1\end{bmatrix}^{\mathsf{T}} =\begin{bmatrix}1\\2\\-1\end{bmatrix}.$$

$$A = \begin{bmatrix}1&2&3\\0&-1&4\\5&6&0\end{bmatrix} \implies A^{\mathsf{T}} = \begin{bmatrix}1&0&5\\2&-1&6\\3&4&0\end{bmatrix}.$$

Notice how the main diagonal ($1,-1,0$) stays fixed.

$$\begin{bmatrix}2&3\\3&5\end{bmatrix}, \qquad \begin{bmatrix}1&0&-2\\0&4&7\\-2&7&0\end{bmatrix}, \qquad \begin{bmatrix}5\end{bmatrix}.$$

In each case, the matrix is its own mirror-image across the main diagonal.

$$A = \begin{bmatrix}0&2&-3\\-2&0&5\\3&-5&0\end{bmatrix}.$$

Check: $a_{12}=2$ and $a_{21}=-2=-a_{12}$; diagonal entries are all $0$. So $A^{\mathsf{T}}=-A$.

Is $A=\begin{bmatrix}1&2&3\\2&0&-1\\3&-1&4\end{bmatrix}$ symmetric?

$A^{\mathsf{T}}=\begin{bmatrix}1&2&3\\2&0&-1\\3&-1&4\end{bmatrix} = A$. Yes, $A$ is symmetric.

$$A=\begin{bmatrix}1&2\\3&4\\5&6\end{bmatrix}, \quad A^{\mathsf{T}}=\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}, \quad (A^{\mathsf{T}})^{\mathsf{T}}=\begin{bmatrix}1&2\\3&4\\5&6\end{bmatrix}=A. \checkmark$$

Let $k=2$, $A=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$.

$$(2A)^{\mathsf{T}}=\begin{bmatrix}2&-2\\4&6\end{bmatrix}^{\mathsf{T}} =\begin{bmatrix}2&4\\-2&6\end{bmatrix}, \quad 2A^{\mathsf{T}}=2\begin{bmatrix}1&2\\-1&3\end{bmatrix} =\begin{bmatrix}2&4\\-2&6\end{bmatrix}. \checkmark$$

$A=\begin{bmatrix}1&0\\2&1\end{bmatrix}$, $B=\begin{bmatrix}3&-1\\0&2\end{bmatrix}$.

$$(A+B)^{\mathsf{T}}=\begin{bmatrix}4&-1\\2&3\end{bmatrix}^{\mathsf{T}} =\begin{bmatrix}4&2\\-1&3\end{bmatrix}.$$

$$A^{\mathsf{T}}+B^{\mathsf{T}}=\begin{bmatrix}1&2\\0&1\end{bmatrix} +\begin{bmatrix}3&0\\-1&2\end{bmatrix} =\begin{bmatrix}4&2\\-1&3\end{bmatrix}. \checkmark$$

Solve for $A$ if $\left(2A^{\mathsf{T}} - 3\begin{bmatrix}1&2\\-1&1\end{bmatrix}\right)^{\mathsf{T}} = \begin{bmatrix}2&3\\-1&2\end{bmatrix}.$

$\textit{Solution.}$ Apply the transpose rules to the left side:

$$\left(2A^{\mathsf{T}} - 3\begin{bmatrix}1&2\\-1&1\end{bmatrix}\right)^{\mathsf{T}} = 2(A^{\mathsf{T}})^{\mathsf{T}} - 3\begin{bmatrix}1&2\\-1&1\end{bmatrix}^{\mathsf{T}} = 2A - 3\begin{bmatrix}1&-1\\2&1\end{bmatrix}.$$

The equation becomes

$$2A - 3\begin{bmatrix}1&-1\\2&1\end{bmatrix} = \begin{bmatrix}2&3\\-1&2\end{bmatrix}.$$

$$2A = \begin{bmatrix}2&3\\-1&2\end{bmatrix} + 3\begin{bmatrix}1&-1\\2&1\end{bmatrix} = \begin{bmatrix}5&0\\5&5\end{bmatrix}.$$

$$A = \dfrac{1}{2}\begin{bmatrix}5&0\\5&5\end{bmatrix} = \begin{bmatrix}\tfrac52&0\\[4pt]\tfrac52&\tfrac52\end{bmatrix}.$$

If $A$ and $B$ are symmetric $n\times n$ matrices, show that $A+B$ is symmetric.

$\textit{Proof.}$ Since $A^{\mathsf{T}}=A$ and $B^{\mathsf{T}}=B$, Theorem 2.1.2(4) gives

$$(A+B)^{\mathsf{T}} = A^{\mathsf{T}} + B^{\mathsf{T}} = A + B.$$

Hence $A+B$ is symmetric.

Let $A$ be a square matrix satisfying $A=2A^{\mathsf{T}}$. Show $A=0$.

$\textit{Proof.}$ Transpose both sides of $A=2A^{\mathsf{T}}$:

$$A^{\mathsf{T}} = (2A^{\mathsf{T}})^{\mathsf{T}} = 2(A^{\mathsf{T}})^{\mathsf{T}} = 2A.$$

Substituting $A^{\mathsf{T}}=2A$ back into $A=2A^{\mathsf{T}}$:

$$A = 2(2A) = 4A.$$

Subtracting $A$ from both sides: $3A=0$, so $A=\tfrac13\cdot 0 = 0$.

$\textbf{(a)}$ $A + B = 3A + 2B$.

Rearrange: $A - 3A = 2B - B$, so $-2A = B$, hence $\boxed{A = -\tfrac{1}{2}B.}$

$\textbf{(b)}$ $2A - B = 5(A + 2B)$.

Expand: $2A - B = 5A + 10B$. Rearrange: $2A - 5A = 10B + B$, so $-3A = 11B$, hence $\boxed{A = -\tfrac{11}{3}B.}$

The system has the same structure as a linear system of two equations in two unknowns, except that the "unknowns" $X$, $Y$ and the "constants" $A$, $B$ are matrices (of the same size). We can therefore apply exactly the same elimination procedure as we use for numbers.

Write the system as an augmented matrix whose entries are the coefficient scalars on the left and the matrix constants on the right:

$$\left[\begin{array}{cc|c} \text{coeff of }X & \text{coeff of }Y & \text{RHS} \end{array}\right].$$

(a) $5X + 3Y = A$ $\quad$ and $\quad$ $2X + Y = B$.

Write the augmented representation and eliminate:

$$\left[\begin{array}{cc|c} 5 & 3 & A \\ 2 & 1 & B \end{array}\right] \xrightarrow{R_1 - 2R_2} \left[\begin{array}{cc|c} 1 & 1 & A-2B \\ 2 & 1 & B \end{array}\right] \xrightarrow{R_2 - 2R_1} \left[\begin{array}{cc|c} 1 & 1 & A-2B \\ 0 & -1 & B-2(A-2B) \end{array}\right]$$

Simplify the bottom-right entry: $B - 2A + 4B = 5B - 2A$.

$$\left[\begin{array}{cc|c} 1 & 1 & A-2B \\ 0 & -1 & 5B-2A \end{array}\right] \xrightarrow{-R_2} \left[\begin{array}{cc|c} 1 & 1 & A-2B \\ 0 & 1 & 2A-5B \end{array}\right] \xrightarrow{R_1 - R_2} \left[\begin{array}{cc|c} 1 & 0 & 3B-A \\ 0 & 1 & 2A-5B \end{array}\right]$$

Reading off directly from the RREF:

$$\boxed{X = 3B - A, \qquad Y = 2A - 5B.}$$

$\textit{Verification:}$

$$5X+3Y = 5(3B-A)+3(2A-5B) = 15B-5A+6A-15B = A.\;\checkmark$$

$$2X+Y = 2(3B-A)+(2A-5B) = 6B-2A+2A-5B = B.\;\checkmark$$

(b) $4X + 3Y = A$ $\quad$ and $\quad$ $5X + 4Y = B$.

$$\left[\begin{array}{cc|c} 4 & 3 & A \\ 5 & 4 & B \end{array}\right] \xrightarrow{4R_2 - 5R_1} \left[\begin{array}{cc|c} 4 & 3 & A \\ 0 & 1 & 4B-5A \end{array}\right] \xrightarrow{R_1 - 3R_2} \left[\begin{array}{cc|c} 4 & 0 & A - 3(4B-5A) \\ 0 & 1 & 4B-5A \end{array}\right]$$

Simplify: $A - 12B + 15A = 16A - 12B$.

$$\left[\begin{array}{cc|c} 4 & 0 & 16A-12B \\ 0 & 1 & 4B-5A \end{array}\right] \xrightarrow{\frac{1}{4}R_1} \left[\begin{array}{cc|c} 1 & 0 & 4A-3B \\ 0 & 1 & -5A+4B \end{array}\right]$$

Reading off:

$$\boxed{X = 4A - 3B, \qquad Y = -5A + 4B.}$$

$\textit{Verification:}$

$$4X+3Y = 4(4A-3B)+3(-5A+4B) = 16A-12B-15A+12B = A.\;\checkmark$$

$$5X+4Y = 5(4A-3B)+4(-5A+4B) = 20A-15B-20A+16B = B.\;\checkmark$$

(a) $2[9(A-B)+7(2B-A)] - 2[3(2B+A)-2(A+3B)-5(A+B)]$.

$\textit{Inner brackets first:}$

$$9(A-B)+7(2B-A) = 9A-9B+14B-7A = 2A+5B.$$

$$3(2B+A)-2(A+3B)-5(A+B) = 6B+3A-2A-6B-5A-5B = -4A-5B.$$

$\textit{Now the full expression:}$

$$2(2A+5B) - 2(-4A-5B) = 4A+10B+8A+10B = \boxed{12A+20B.}$$

(b) $5[3(A-B+2C)-2(3C-B)-A]+2[3(3A-B+C)+2(B-2A)-2C]$.

$\textit{Inner brackets:}$

$$3(A-B+2C)-2(3C-B)-A = 3A-3B+6C-6C+2B-A = 2A-B.$$

$$3(3A-B+C)+2(B-2A)-2C = 9A-3B+3C+2B-4A-2C = 5A-B+C.$$

$\textit{Full expression:}$

$$5(2A-B)+2(5A-B+C) = 10A-5B+10A-2B+2C = \boxed{20A-7B+2C.}$$

$\textbf{(a)}$ Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ be any $2\times2$ matrix. Then:

$$A = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}.$$

$\textit{Proof.}$ Add the four scalar multiples entry by entry:

$$\begin{bmatrix}a&0\\0&0\end{bmatrix} +\begin{bmatrix}0&b\\0&0\end{bmatrix} +\begin{bmatrix}0&0\\c&0\end{bmatrix} +\begin{bmatrix}0&0\\0&d\end{bmatrix} =\begin{bmatrix}a&b\\c&d\end{bmatrix}=A.$$

Setting $a,b,c,d$ equal to the entries of $A$ gives the required representation.

$\textbf{(b)}$ We claim $A = p\begin{bmatrix}1&0\\0&1\end{bmatrix} +q\begin{bmatrix}1&1\\0&0\end{bmatrix} +r\begin{bmatrix}1&0\\1&0\end{bmatrix} +s\begin{bmatrix}0&1\\1&0\end{bmatrix}$ for some $p,q,r,s$.

Expanding:

$$\begin{bmatrix}p+q+r & q+s \\ r+s & p\end{bmatrix} = \begin{bmatrix}a&b\\c&d\end{bmatrix}.$$

Matching entries: $p=d$, $r+s=c$, $q+s=b$, $p+q+r=a$. From $p=d$: $q+r = a-d$. From $q+s=b$ and $r+s=c$: subtracting gives $q-r=b-c$, so $q=\tfrac{(a-d)+(b-c)}{2}$, $r=\tfrac{(a-d)-(b-c)}{2}$, $s=b-q$. This always has a solution, so the representation exists for any $A$.

Given $A=[1\;1\;{-1}]$, $B=[0\;1\;2]$, $C=[3\;0\;1]$, and $rA+sB+tC=0$ for scalars $r,s,t$. Writing this out entry by entry:

$$\text{entry 1:}\; r + 0 + 3t = 0,\quad \text{entry 2:}\; r + s + 0 = 0,\quad \text{entry 3:}\; -r + 2s + t = 0.$$

This is a homogeneous system in $r,s,t$:

$$\begin{bmatrix}1&0&3\\1&1&0\\-1&2&1\end{bmatrix} \begin{pmatrix}r\\s\\t\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$

Computing the determinant: $1(1\cdot1-0\cdot2)-0+3(1\cdot2-1\cdot(-1)) = 1+3(3)=10\neq0$. Since the coefficient matrix is nonsingular, the only solution is $r=s=t=0$.

$\textbf{(a)}$ Suppose $Q+A=A$ holds for every $m\times n$ matrix $A$. Show $Q=O_{mn}$.

$\textit{Proof.}$ By hypothesis, $Q+A=A$. Subtracting $A$: $Q=A-A=O$.

$\textbf{(b)}$ Suppose $A+A'=O_{mn}$. Show $A'=-A$.

$\textit{Proof.}$ $A+A'=O$ means $A'=O-A=-A$.

$\textbf{($\Rightarrow$)}$ Assume $A=-A$. Then $A+A=0$, so $2A=O$, hence $A=\tfrac{1}{2}O=O$.

$\textbf{($\Leftarrow$)}$ If $A=O$, then $-A=-O=O=A$.

Therefore $A=-A$ if and only if $A=O$.

Let $A=[a_{ij}]$ and $B=[b_{ij}]$ be diagonal $n\times n$ matrices, so $a_{ij}=0$ and $b_{ij}=0$ for all $i\neq j$.

$\textbf{(a)}$ $A+B$ is diagonal. The $(i,j)$-entry of $A+B$ is $a_{ij}+b_{ij}$. For $i\neq j$: $a_{ij}+b_{ij}=0+0=0$. So all off-diagonal entries of $A+B$ are zero; hence $A+B$ is diagonal.

$\textbf{(b)}$ $A-B$ is diagonal. The $(i,j)$-entry of $A-B$ is $a_{ij}-b_{ij}$. For $i\neq j$: $a_{ij}-b_{ij}=0-0=0$. So $A-B$ is diagonal.

$\textbf{(c)}$ $kA$ is diagonal for any scalar $k$. The $(i,j)$-entry of $kA$ is $k\,a_{ij}$. For $i\neq j$: $k\cdot a_{ij}=k\cdot0=0$. So $kA$ is diagonal.

A matrix is symmetric iff $a_{ij}=a_{ji}$ for all $i\neq j$.

$\textbf{(a)}$ $\begin{bmatrix}1&s\\-2&t\end{bmatrix}$ symmetric requires $a_{12}=a_{21}$, i.e. $s=-2$. No condition on $t$ (it is a diagonal entry). Answer: $\boxed{s=-2,\; t\in\mathbb{R}}$ (any $t$).

$\textbf{(b)}$ $\begin{bmatrix}s&t\\st&1\end{bmatrix}$ symmetric requires $a_{12}=a_{21}$, i.e. $t=st$. $t=st \Rightarrow t(s-1)=0 \Rightarrow t=0$ or $s=1$.

$$\boxed{t=0 \text{ (any }s),\quad\text{or}\quad s=1 \text{ (any }t).}$$

$\textbf{(c)}$ $\begin{bmatrix}s&2s&st\\t&-1&s\\t&s^2&s\end{bmatrix}$ symmetric requires:

$$\begin{aligned} a_{12}&=a_{21}: & 2s&=t,\\ a_{13}&=a_{31}: & st&=t,\\ a_{23}&=a_{32}: & s&=s^2. \end{aligned}$$

From $s=s^2$: $s(s-1)=0$, so $s=0$ or $s=1$.

• $s=0$: $t=2(0)=0$. Check $st=0=t$. $\checkmark$ $\boxed{s=0,\;t=0.}$
• $s=1$: $t=2(1)=2$. Check $st=2=t$. $\checkmark$ $\boxed{s=1,\;t=2.}$

$\textbf{(d)}$ $\begin{bmatrix}2&s&t\\2s&0&s+t\\3&3&t\end{bmatrix}$ symmetric requires:

$$\begin{aligned} a_{12}&=a_{21}: & s&=2s \implies s=0,\\ a_{13}&=a_{31}: & t&=3,\\ a_{23}&=a_{32}: & s+t&=3 \implies 0+3=3.\checkmark \end{aligned}$$

$$\boxed{s=0,\; t=3.}$$

$\textbf{(a)}$ $\left(A+3\begin{bmatrix}1&-1&0\\1&2&4\end{bmatrix}\right)^{\mathsf{T}} =\begin{bmatrix}2&1\\0&5\\3&8\end{bmatrix}$.

Transpose both sides using $(P+Q)^{\mathsf{T}}=P^{\mathsf{T}}+Q^{\mathsf{T}}$ and $(M^{\mathsf{T}})^{\mathsf{T}}=M$:

$$A + 3\begin{bmatrix}1&-1&0\\1&2&4\end{bmatrix} = \begin{bmatrix}2&1\\0&5\\3&8\end{bmatrix}^{\mathsf{T}} = \begin{bmatrix}2&0&3\\1&5&8\end{bmatrix}.$$

$$A = \begin{bmatrix}2&0&3\\1&5&8\end{bmatrix} - 3\begin{bmatrix}1&-1&0\\1&2&4\end{bmatrix} = \begin{bmatrix}2-3&0+3&3-0\\1-3&5-6&8-12\end{bmatrix} = \boxed{\begin{bmatrix}-1&3&3\\-2&-1&-4\end{bmatrix}.}$$

$\textbf{(b)}$ $\left(3A^{\mathsf{T}}+2\begin{bmatrix}1&0\\0&2\end{bmatrix}\right)^{\mathsf{T}} =\begin{bmatrix}8&0\\3&1\end{bmatrix}$.

Apply $(P+Q)^{\mathsf{T}}=P^{\mathsf{T}}+Q^{\mathsf{T}}$ and $(kA)^{\mathsf{T}}=kA^{\mathsf{T}}$:

$$3(A^{\mathsf{T}})^{\mathsf{T}}+2\begin{bmatrix}1&0\\0&2\end{bmatrix}^{\mathsf{T}} =\begin{bmatrix}8&0\\3&1\end{bmatrix} \implies 3A+2\begin{bmatrix}1&0\\0&2\end{bmatrix} =\begin{bmatrix}8&0\\3&1\end{bmatrix}.$$

$$3A=\begin{bmatrix}8&0\\3&1\end{bmatrix} -\begin{bmatrix}2&0\\0&4\end{bmatrix} =\begin{bmatrix}6&0\\3&-3\end{bmatrix}.$$

$$A=\tfrac{1}{3}\begin{bmatrix}6&0\\3&-3\end{bmatrix} =\boxed{\begin{bmatrix}2&0\\1&-1\end{bmatrix}.}$$

$\textbf{(c)}$ $(2A-3[1\;2\;0])^{\mathsf{T}} = 3A^{\mathsf{T}}+[2\;1\;{-1}]^{\mathsf{T}}$.

Apply $(\cdot)^{\mathsf{T}}$ rules to the left side. Since both sides involve $A^{\mathsf{T}}$, note $[1\;2\;0]$ is $1\times3$, so $A$ must be $1\times3$. Transposing the left side:

$$(2A-3[1\;2\;0])^{\mathsf{T}} = 2A^{\mathsf{T}}-3[1\;2\;0]^{\mathsf{T}} = 2A^{\mathsf{T}}-3\begin{bmatrix}1\\2\\0\end{bmatrix}.$$

The equation is then:

$$2A^{\mathsf{T}}-3\begin{bmatrix}1\\2\\0\end{bmatrix} = 3A^{\mathsf{T}}+\begin{bmatrix}2\\1\\-1\end{bmatrix}.$$

$$2A^{\mathsf{T}}-3A^{\mathsf{T}} = \begin{bmatrix}2\\1\\-1\end{bmatrix}+\begin{bmatrix}3\\6\\0\end{bmatrix} \implies -A^{\mathsf{T}} = \begin{bmatrix}5\\7\\-1\end{bmatrix} \implies A^{\mathsf{T}} = \begin{bmatrix}-5\\-7\\1\end{bmatrix}.$$

$$A = (A^{\mathsf{T}})^{\mathsf{T}} = \boxed{[-5\;-7\;1].}$$

$\textbf{(d)}$ $\left(2A^{\mathsf{T}}-5\begin{bmatrix}1&0\\-1&2\end{bmatrix}\right)^{\mathsf{T}} = 4A-9\begin{bmatrix}1&1\\-1&0\end{bmatrix}$.

Transpose the left side:

$$2(A^{\mathsf{T}})^{\mathsf{T}} - 5\begin{bmatrix}1&0\\-1&2\end{bmatrix}^{\mathsf{T}} = 2A-5\begin{bmatrix}1&-1\\0&2\end{bmatrix}.$$

The equation becomes:

$$2A-5\begin{bmatrix}1&-1\\0&2\end{bmatrix} =4A-9\begin{bmatrix}1&1\\-1&0\end{bmatrix}.$$

$$2A-4A = -9\begin{bmatrix}1&1\\-1&0\end{bmatrix} +5\begin{bmatrix}1&-1\\0&2\end{bmatrix} = \begin{bmatrix}-9&-9\\9&0\end{bmatrix}+\begin{bmatrix}5&-5\\0&10\end{bmatrix} = \begin{bmatrix}-4&-14\\9&10\end{bmatrix}.$$

$$-2A=\begin{bmatrix}-4&-14\\9&10\end{bmatrix} \implies A=\boxed{\begin{bmatrix}2&7\\-\frac{9}{2}&-5\end{bmatrix}.}$$

We can now add matrices, scale them, and flip them. But how do we multiply two matrices?

Matrix multiplication is the operation that makes linear algebra powerful — and its definition is far less obvious than addition. Next lecture we build matrix–vector and matrix–matrix products, discover why multiplication is not commutative, and connect it back to the linear systems we have been solving all along.