Matrices and Matrix Operations — Worked Solutions

Ques­tion 1: Build­ing a matrix from a rule#

The prob­lem#

Con­struct the $3 \times 2$ matrix with $a_{ij} = \lvert 2i - j \rvert$.

Under­stand­ing the prob­lem#

The matrix has $3$ rows ($i = 1, 2, 3$) and $2$ columns ($j = 1, 2$). The entry in row $i$ and col­umn $j$ is $\lvert 2i - j \rvert$. You must work out all six entries.

The idea#

Go row by row, sub­sti­tut­ing each pair $(i, j)$ into the rule.

Step-by-step solu­tion#

Step 1. Row 1 ($i = 1$).

$$a_{11} = \lvert 2 - 1 \rvert = 1, \qquad a_{12} = \lvert 2 - 2 \rvert = 0.$$

Step 2. Row 2 ($i = 2$).

$$a_{21} = \lvert 4 - 1 \rvert = 3, \qquad a_{22} = \lvert 4 - 2 \rvert = 2.$$

Step 3. Row 3 ($i = 3$).

$$a_{31} = \lvert 6 - 1 \rvert = 5, \qquad a_{32} = \lvert 6 - 2 \rvert = 4.$$

Step 4. Arrange them.

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

Check­ing the answer#

The order is $3 \times 2$ (three rows, two columns) ✓. Every entry is non-neg­a­tive, as a mod­u­lus must be ✓.

Answer#

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

Com­mon mis­take to avoid#

$i$ is the row num­ber and $j$ the col­umn num­ber. Swap­ping them gives a $2 \times 3$ matrix with the wrong entries.

Ques­tion 2: Unknowns from equal matri­ces#

The prob­lem#

Find $x, y, z$: $\begin{bmatrix} x + y & 2 \\ 5 & xz \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 5 & 12 \end{bmatrix}$ with $x = 4$.

Under­stand­ing the prob­lem#

Two matri­ces are equal when every entry matches its part­ner in the same posi­tion. You are also told $x = 4$.

The idea#

Equate cor­re­spond­ing entries to get equa­tions, then sub­sti­tute $x = 4$.

Step-by-step solu­tion#

Step 1. Com­pare entries: $(1,1)$ gives $x + y = 7$; $(2,2)$ gives $xz = 12$. The other entries ($2 = 2$, $5 = 5$) already match.

Step 2. Put $x = 4$ in the first equa­tion.

$$4 + y = 7 \quad\Longrightarrow\quad y = 3.$$

Step 3. Put $x = 4$ in the sec­ond.

$$4z = 12 \quad\Longrightarrow\quad z = 3.$$

Check­ing the answer#

$x + y = 7$ ✓ and $xz = 12$ ✓.

Answer#

$x = 4$, $y = 3$, $z = 3$

Ques­tion 3: A lin­ear com­bi­na­tion of matri­ces#

The prob­lem#

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

Under­stand­ing the prob­lem#

$A$ and $B$ have the same order ($2 \times 3$), so they can be com­bined. You need to mul­ti­ply each by a num­ber and sub­tract.

The idea#

Scalar mul­ti­pli­ca­tion mul­ti­plies every entry. Then sub­tract entry by entry.

Step-by-step solu­tion#

Step 1. Mul­ti­ply $A$ by $2$.

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

Step 2. Mul­ti­ply $B$ by $3$.

$$3B = \begin{bmatrix} 6 & 0 & 3 \\ -3 & 9 & 15 \end{bmatrix}.$$

Step 3. Sub­tract posi­tion by posi­tion.

$$2A - 3B = \begin{bmatrix} 2 - 6 & 4 - 0 & -6 - 3 \\ 0 - (-3) & 8 - 9 & 2 - 15 \end{bmatrix} = \begin{bmatrix} -4 & 4 & -9 \\ 3 & -1 & -13 \end{bmatrix}.$$

Check­ing the answer#

Check one entry directly: posi­tion $(2, 3)$ is $2(1) - 3(5) = 2 - 15 = -13$ ✓.

Answer#

$2A - 3B = \begin{bmatrix} -4 & 4 & -9 \\ 3 & -1 & -13 \end{bmatrix}$

Ques­tion 4: Solv­ing a matrix equa­tion#

The prob­lem#

Find $X$ if $3X + \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 7 & -4 \\ 9 & 5 \end{bmatrix}$.

Under­stand­ing the prob­lem#

$X$ is an unknown $2 \times 2$ matrix. Treat the equa­tion like an ordi­nary lin­ear equa­tion, but with matrix addi­tion and scalar mul­ti­pli­ca­tion.

The idea#

Sub­tract the known matrix from both sides, then divide every entry by $3$.

Step-by-step solu­tion#

Step 1. Sub­tract $\begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$ from both sides.

$$3X = \begin{bmatrix} 7 - 1 & -4 - 2 \\ 9 - 0 & 5 - (-1) \end{bmatrix} = \begin{bmatrix} 6 & -6 \\ 9 & 6 \end{bmatrix}.$$

Step 2. Mul­ti­ply by $\displaystyle \tfrac{1}{3}$.

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

Check­ing the answer#

$3X = \begin{bmatrix} 6 & -6 \\ 9 & 6 \end{bmatrix}$; adding $\begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$ gives $\begin{bmatrix} 7 & -4 \\ 9 & 5 \end{bmatrix}$ ✓.

Answer#

$X = \begin{bmatrix} 2 & -2 \\ 3 & 2 \end{bmatrix}$

Ques­tion 5: Two matrix prod­ucts#

The prob­lem#

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

Under­stand­ing the prob­lem#

First check the prod­ucts are defined. (a) $2 \times 2$ times $2 \times 2$ gives $2 \times 2$. (b) $1 \times 3$ times $3 \times 1$ gives $1 \times 1$.

The idea#

Each entry $(i, k)$ of the prod­uct is row $i$ of the first matrix times col­umn $k$ of the sec­ond: mul­ti­ply match­ing entries and add.

Step-by-step solu­tion#

Part (a)

Step 1. Row 1 $(1, 3)$ with col­umn 1 $(2, 4)$ and col­umn 2 $(-1, 5)$.

$$1(2) + 3(4) = 14, \qquad 1(-1) + 3(5) = 14.$$

Step 2. Row 2 $(-2, 0)$ with the same columns.

$$-2(2) + 0(4) = -4, \qquad -2(-1) + 0(5) = 2.$$

Step 3. Assem­ble.

$$\begin{bmatrix} 14 & 14 \\ -4 & 2 \end{bmatrix}.$$

Part (b)

Step 1. One row times one col­umn gives one entry.

$$2(3) + 1(-1) + 0(7) = 6 - 1 + 0 = 5.$$

So the prod­uct is the $1 \times 1$ matrix $[5]$.

Check­ing the answer#

The orders match the pre­dic­tions ($2 \times 2$ and $1 \times 1$) ✓.

Answer#

$\begin{bmatrix} 14 & 14 \\ -4 & 2 \end{bmatrix}$; $[5]$.

Ques­tion 6: A matrix sat­is­fy­ing a qua­dratic equa­tion#

The prob­lem#

For $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$, show $A^2 - 4A + I = O$.

Under­stand­ing the prob­lem#

You must com­pute $A^2 = AA$, then $4A$ and $I$, com­bine them, and show every entry is $0$.

The idea#

Work out $A^2$ by row-into-col­umn mul­ti­pli­ca­tion, then check each of the four posi­tions.

Step-by-step solu­tion#

Step 1. Com­pute $A^2$.

$$A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}.$$

Step 2. Com­pute $4A$ and write $I$.

$$4A = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix}, \qquad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Step 3. Com­bine entry by entry.

$$A^2 - 4A + I = \begin{bmatrix} 7 - 8 + 1 & 12 - 12 + 0 \\ 4 - 4 + 0 & 7 - 8 + 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O.$$

Check­ing the answer#

All four entries are $0$, which is exactly what had to be shown ✓.

Answer#

$A^2 = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$, and $A^2 - 4A + I = O$ as required.

Com­mon mis­take to avoid#

$A^2$ is not found by squar­ing each entry. You must mul­ti­ply the matrix by itself.

Ques­tion 7: Ver­i­fy­ing the rever­sal law for trans­poses#

The prob­lem#

Ver­ify $(AB)^T = B^TA^T$ for $A = \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 1 \end{bmatrix}$.

Under­stand­ing the prob­lem#

"Ver­ify" means com­pute both sides and show they are equal. $A$ is $3 \times 1$ and $B$ is $1 \times 2$, so $AB$ is $3 \times 2$ and $(AB)^T$ is $2 \times 3$.

The idea#

Find $AB$, trans­pose it. Sep­a­rately, find $B^T$ and $A^T$ and mul­ti­ply them in that order.

Step-by-step solu­tion#

Step 1. Com­pute $AB$: each row of $A$ has one entry, mul­ti­plied by each entry of $B$.

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

Step 2. Trans­pose (rows become columns).

$$(AB)^T = \begin{bmatrix} 2 & -4 & 6 \\ 1 & -2 & 3 \end{bmatrix}.$$

Step 3. Write $B^T$ ($2 \times 1$) and $A^T$ ($1 \times 3$).

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

Step 4. Mul­ti­ply $B^TA^T$ ($2 \times 1$ times $1 \times 3$ gives $2 \times 3$).

$$B^TA^T = \begin{bmatrix} 2(1) & 2(-2) & 2(3) \\ 1(1) & 1(-2) & 1(3) \end{bmatrix} = \begin{bmatrix} 2 & -4 & 6 \\ 1 & -2 & 3 \end{bmatrix}.$$

Step 5. The results of Steps 2 and 4 are iden­ti­cal, so $(AB)^T = B^TA^T$.

Check­ing the answer#

Note that $A^TB^T$ would be $1 \times 3$ times $2 \times 1$, which is not even defined — that is why the order must reverse.

Answer#

Both sides equal $\begin{bmatrix} 2 & -4 & 6 \\ 1 & -2 & 3 \end{bmatrix}$, so $(AB)^T = B^TA^T$ holds.

Ques­tion 8: Orders of prod­ucts#

The prob­lem#

If $A$ is $3 \times 4$ and $B$ is $4 \times 2$, what is the order of $AB$? Is $BA$ defined?

Under­stand­ing the prob­lem#

A prod­uct $PQ$ is defined when (columns of $P$) = (rows of $Q$). Its order is (rows of $P$) × (columns of $Q$).

The idea#

Write the orders side by side and check the inner num­bers.

Step-by-step solu­tion#

Step 1. For $AB$: $(3 \times \mathbf{4})(\mathbf{4} \times 2)$. The inner num­bers match ($4 = 4$), so $AB$ is defined, with order given by the outer num­bers:

$$3 \times 2.$$

Step 2. For $BA$: $(4 \times \mathbf{2})(\mathbf{3} \times 4)$. The inner num­bers are $2$ and $3$, which dif­fer, so $BA$ is not defined.

Check­ing the answer#

Each row of $B$ has $2$ entries but each col­umn of $A$ has $3$, so there is noth­ing to pair up — con­firm­ing $BA$ can­not be formed.

Answer#

$AB$ is $3 \times 2$; $BA$ is not defined.

Ques­tion 9: Cost of two items using matri­ces#

The prob­lem#

A fac­tory makes items X and Y using $2$ and $3$ hours of labour and $4$ and $1$ kg of mate­r­ial. Labour costs ₹150 per hour and mate­r­ial ₹60 per kg. Use matri­ces to find the cost of each item.

Under­stand­ing the prob­lem#

Item X needs $2$ hours of labour and $4$ kg of mate­r­ial. Item Y needs $3$ hours and $1$ kg. The cost of each item is (hours × ₹150) + (kg × ₹60).

The idea#

Put the require­ments in a matrix with one row per item and one col­umn per resource, and the unit costs in a col­umn matrix. Their prod­uct gives the cost of each item.

Step-by-step solu­tion#

Step 1. Require­ments matrix (rows: X, Y; columns: labour, mate­r­ial).

$$R = \begin{bmatrix} 2 & 4 \\ 3 & 1 \end{bmatrix}.$$

Step 2. Cost col­umn (labour, mate­r­ial).

$$C = \begin{bmatrix} 150 \\ 60 \end{bmatrix}.$$

Step 3. Mul­ti­ply ($2 \times 2$ times $2 \times 1$ gives $2 \times 1$).

$$RC = \begin{bmatrix} 2(150) + 4(60) \\ 3(150) + 1(60) \end{bmatrix} = \begin{bmatrix} 300 + 240 \\ 450 + 60 \end{bmatrix} = \begin{bmatrix} 540 \\ 510 \end{bmatrix}.$$

Check­ing the answer#

X: 2 hours cost ₹300 and 4 kg cost ₹240, total ₹540 ✓. Y: ₹450 + ₹60 = ₹510 ✓.

Answer#

Item X costs ₹540 and item Y costs ₹510.

Ques­tion 10: Two non-zero matri­ces with zero prod­uct#

The prob­lem#

Find two non-zero $2 \times 2$ matri­ces whose prod­uct is $O$.

Under­stand­ing the prob­lem#

With num­bers, $ab = 0$ forces $a = 0$ or $b = 0$. With matri­ces this is not true. You need one exam­ple of $A \ne O$, $B \ne O$ with $AB = O$.

The idea#

Choose $A$ that keeps only the first "slot" and $B$ that has some­thing only in the sec­ond slot, so every row of $A$ meets a col­umn of $B$ with no over­lap­ping non-zero entries.

Step-by-step solu­tion#

Step 1. Take

$$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$$

Both are non-zero (each has an entry $1$).

Step 2. Mul­ti­ply.

$$AB = \begin{bmatrix} 1(0) + 0(0) & 1(0) + 0(1) \\ 0(0) + 0(0) & 0(0) + 0(1) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O.$$

Check­ing the answer#

Every entry of $AB$ is $0$ ✓. The lesson's own exam­ple, $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = O$, is another cor­rect answer; many oth­ers exist.

Answer#