Matrices and Matrix Operations: Precalculus Math Worksheet Answer Key – PDF Download

Full worked solutions for all 23 problems follow, with each matrix drawn and every row-by-column sum shown. Check your own entries against the drawn results, and verify any inverse by multiplying it back to the identity.

PrecalculusCommon Core: N-VM.C.6 · N-VM.C.7 · N-VM.C.8 · N-VM.C.9 · N-VM.C.10

Problem 1: Dimensions and entries

$G$ has 3 rows and 4 columns, so its dimensions are $3\times 4$.

Row 2, column 3: $-3$
Row 3, column 4: $-2$
The entry 7 is in row 1, column 3.

Problem 2: Organize data in a matrix

The matrix has 3 rows and 2 columns, so it is $3\times 2$.

Bake sale data as a 3 × 2 matrix.

Muffins are row 3: $30+21=$ 51 muffins.

Problem 3: Add matrices

Add matching entries: $3+(-1)=2$, $-2+4=2$, $5+2=7$, $0+6=6$.

The sum A + B.

Problem 4: Subtract matrices

Subtract matching entries: row 1 is $7-4=3$, $2-(-1)=3$, $-3-2=-5$; row 2 is $0-3=-3$, $5-5=0$, $1-(-6)=7$.

The difference C − D.

Problem 5: Scalar multiples

Multiply every entry by the scalar.

The scalar multiples 3A and −2B.

Problem 6: Combine operations

First $2A$ has entries 6, $-4$, 10, 0. Then $3B$ has entries $-3$, 12, 6, 18. Subtract entry by entry: $6-(-3)=9$, $-4-12=-16$, $10-6=4$, $0-18=-18$.

The result 2A − 3B.

Problem 7: Solve for unknowns

Equal matrices have equal entries. Top left: $x+3=7$, so $x=4$. Top right: $2+y=5$, so $y=3$.

Check with the bottom row: $5+x=9$ and $y+1=4$. Both hold, so $x=4$ and $y=3$.

Problem 8: Is the product defined?

Columns of $A$ = 3 = rows of $B$: defined, $2\times 4$
Columns of $A$ = 2 but rows of $B$ = 3: not defined
Inner numbers match (3 and 3): defined, $1\times 1$
Inner numbers match (1 and 1): defined, $3\times 3$

Problem 9: Multiply 2 × 2 matrices

$c_{11}=1\cdot 4+3\cdot (-2)=-2$
$c_{12}=1\cdot 0+3\cdot 5=15$
$c_{21}=2\cdot 4+(-1)\cdot (-2)=10$
$c_{22}=2\cdot 0+(-1)\cdot 5=-5$

The product PQ.

Problem 10: Multiply a 2 × 3 by a 3 × 2

$R$ is $2\times 3$ and $S$ is $3\times 2$, so $RS$ is $2\times 2$.

$c_{11}=1\cdot 2+0\cdot 0+2\cdot 4=10$
$c_{12}=1\cdot 1+0\cdot (-2)+2\cdot 3=7$
$c_{21}=(-1)\cdot 2+3\cdot 0+1\cdot 4=2$
$c_{22}=(-1)\cdot 1+3\cdot (-2)+1\cdot 3=-4$

The product RS.

Problem 11: Matrix times a column

The result is $2\times 1$. Row 1: $2\cdot 5+(-1)\cdot 2=8$. Row 2: $3\cdot 5+4\cdot 2=23$.

The product, a 2 × 1 column.

Problem 12: Compare AB and BA

$AB$: $c_{11}=1\cdot 3+2\cdot 1=5$, $c_{12}=1\cdot 0+2\cdot 1=2$, $c_{21}=0\cdot 3+1\cdot 1=1$, $c_{22}=0\cdot 0+1\cdot 1=1$.

$BA$: $d_{11}=3\cdot 1+0\cdot 0=3$, $d_{12}=3\cdot 2+0\cdot 1=6$, $d_{21}=1\cdot 1+1\cdot 0=1$, $d_{22}=1\cdot 2+1\cdot 1=3$.

The two products.

The matrices differ, so $AB\neq BA$.

Problem 13: Identity and zero matrices

Multiplying by the identity changes nothing: $AI=A$, the matrix with rows 4, $-3$ and 2, 7.
Adding the zero matrix changes nothing: $A+O=A$.
Every row-by-column sum is 0: $AO=O$, the $2\times 2$ zero matrix.
$-A$ has entries $-4,3,-2,-7$, so the sum is $O$.

Problem 14: Check the distributive property

Left side: $B+C$ has rows 3, 0 and $-1$, 3. Then $A(B+C)$:

$c_{11}=1\cdot 3+2\cdot (-1)=1$
$c_{12}=1\cdot 0+2\cdot 3=6$
$c_{21}=3\cdot 3+0\cdot (-1)=9$
$c_{22}=3\cdot 0+0\cdot 3=0$

Right side: $AB$ has rows 2, 1 and 6, $-3$; $AC$ has rows $-1$, 5 and 3, 3. Adding gives rows 1, 6 and 9, 0.

Both sides agree.

The two sides are equal, so $A(B+C)=AB+AC$.

Problem 15: Multiply 3 × 3 matrices

Each entry is a row of $X$ times a column of $Y$. Row 1 of the product: $1\cdot 2+0\cdot 1+2\cdot 0=2$, $1\cdot 1+0\cdot 0+2\cdot 2=5$, $1\cdot 0+0\cdot 1+2\cdot 1=2$. Row 2: $1$, $-2$, $0$. Row 3: $7$, $3$, $1$.

The product XY.

Problem 16: Find determinants

$3\cdot 4-5\cdot 2=12-10=$ 2
$(-2)(-3)-6\cdot 1=6-6=$ 0
$0\cdot 7-4\cdot(-3)=0+12=$ 12
$5\cdot 3-(-1)\cdot 2=15+2=$ 17

Problem 17: Find k for a singular matrix

$\det=3k-12$. Set $3k-12=0$: $k=4$.
$\det=k^{2}-16$. Set $k^{2}-16=0$: $k=4$ or $k=-4$.

Problem 18: Find a 2 × 2 inverse

$\det=3\cdot 2-1\cdot 5=1$. Swap 3 and 2, negate 1 and 5, and multiply by $\frac{1}{1}$: the inverse has rows 2, $-1$ and $-5$, 3.
$\det=4\cdot 2-2\cdot 3=2$. The swapped matrix has rows 2, $-2$ and $-3$, 4. Multiply by $\frac{1}{2}$: rows 1, $-1$ and $-\frac{3}{2}$, 2.

The two inverses.

Problem 19: Verify an inverse

$AB$: $c_{11}=3\cdot 2+1\cdot (-5)=1$, $c_{12}=3\cdot (-1)+1\cdot 3=0$, $c_{21}=5\cdot 2+2\cdot (-5)=0$, $c_{22}=5\cdot (-1)+2\cdot 3=1$.

$BA$: $d_{11}=2\cdot 3+(-1)\cdot 5=1$, $d_{12}=2\cdot 1+(-1)\cdot 2=0$, $d_{21}=(-5)\cdot 3+3\cdot 5=0$, $d_{22}=(-5)\cdot 1+3\cdot 2=1$.

Both products are the identity matrix.

Since $AB=BA=I$, $B=A^{-1}$.

Problem 20: School store revenue

$S$ is $2\times 3$ and $P$ is $3\times 1$, so $SP$ is $2\times 1$.

Monday: $12(0.50)+8(0.75)+5(2.25)=6+6+11.25=23.25$
Tuesday: $15(0.50)+6(0.75)+9(2.25)=7.5+4.5+20.25=32.25$

Revenue by day.

Revenue was \$23.25 on Monday and \$32.25 on Tuesday.

Problem 21: Farmers market inventory

Add matching entries, then multiply by 2. For example, Saturday apples: $40+10=50$, and doubled $2\cdot 50=100$.

Totals now and doubled totals.

Problem 22: Encode and decode a message

Coded: row 1 is $2\cdot 7+1\cdot 12=26$ and row 2 is $3\cdot 7+2\cdot 12=45$.

The coded column.

$\det K=2\cdot 2-1\cdot 3=1$, so $K^{-1}$ has rows 2, $-1$ and $-3$, 2. Decode: $2\cdot 26+(-1)\cdot 45=7$ and $(-3)\cdot 26+2\cdot 45=12$. The original message is $(7,12)$.

Problem 23: Does the matrix have an inverse?

$\det N=6\cdot 6-9\cdot 4=36-36=0$. A 2 × 2 matrix has an inverse only when its determinant is not zero, so Noah is wrong: $N$ has no inverse. Positive entries do not guarantee a nonzero determinant.

Keep practicing

See every chapter on the Precalculus math page.

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