Solving Systems with Matrices: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Matrix Systems Worksheet Answer Key: problems 1–5

Problem 1: Write a system as AX = B

Rows of \(A\) hold the coefficients of each equation, so \(A\) has rows \(3,\ 1\) and \(2,\ -1\). The unknown column is \(X=(x,\ y)\) and the constant column is \(B=(7,\ 3)\).

Check by multiplying: row 1 gives \(3x+y=7\) and row 2 gives \(2x-y=3\). \(AX=B\) with the matrices shown

Problem 2: Read a matrix equation

Row 1 times \(X\) gives \(4x-y=9\). Row 2 gives \(2x+5y=-1\).

Test \((2,\ -1)\): \(4(2)-(-1)=8+1=9\) and \(2(2)+5(-1)=4-5=-1\). Both hold. The system is \(4x-y=9\), \(2x+5y=-1\), and \((2,\ -1)\) is a solution.

Problem 3: Determinants of 2 by 2 matrices

Use \(ad-bc\).

1. \(3(4)-5(2)=12-10=2\)
2. \((-2)(-3)-6(1)=6-6=0\)
3. \(4(2)-(-1)(3)=8+3=11\)
4. \(\frac{1}{2}(8)-3(2)=4-6=-2\)

(a) 2 (b) 0 (c) 11 (d) \(-2\)

Problem 4: Does the inverse exist?

1. \(6(3)-9(2)=0\): no inverse.
2. \(5(1)-2(3)=-1\): inverse exists.
3. \((-4)(-2)-8(1)=0\): no inverse.
4. \(7(3)-1(0)=21\): inverse exists.

Inverses exist for (b) and (d) only.

Problem 5: Inverse with determinant 1

\(\det=3(2)-5(1)=1\). Swap the diagonal entries and negate the others:

\(2\)	\(-5\)
\(-1\)	\(3\)

The inverse has rows \(2,\ -5\) and \(-1,\ 3\).

Precalculus Matrix Systems Worksheet Answer Key: problems 6–10

Problem 6: Inverses and the identity

1. \(\det=8-6=2\), so \(A^{-1}=\frac{1}{2}\) times the matrix with rows \(2,\ -3\) and \(-2,\ 4\), which is:
   

   \(1\)	\(-\frac{3}{2}\)
   \(-1\)	\(2\)

   Check: first row of \(A\) times first column of \(A^{-1}\) is \(4(1)+3(-1)=1\); first row times second column is \(4(-\frac{3}{2})+3(2)=0\); second row gives \(2(1)+2(-1)=0\) and \(2(-\frac{3}{2})+2(2)=1\).

2. \(\det=20+6=26\), so \(A^{-1}=\frac{1}{26}\) times rows \(4,\ -2\) and \(3,\ 5\):
   

   \(\frac{2}{13}\)	\(-\frac{1}{13}\)
   \(\frac{3}{26}\)	\(\frac{5}{26}\)

3. \(\det=5-6=-1\), so \(A^{-1}\) has rows \(-5,\ 2\) and \(3,\ -1\):
   

   \(-5\)	\(2\)
   \(3\)	\(-1\)

See the three inverses above.

Problem 7: Are these inverses?

\(AC\): row 1 gives \(2(2)+3(-1)=1\) and \(2(-3)+3(2)=0\); row 2 gives \(1(2)+2(-1)=0\) and \(1(-3)+2(2)=1\). So \(AC=I\).

\(CA\): row 1 gives \(2(2)-3(1)=1\) and \(2(3)-3(2)=0\); row 2 gives \(-1(2)+2(1)=0\) and \(-1(3)+2(2)=1\). So \(CA=I\).

Yes, \(C=A^{-1}\).

Problem 8: Solve with a given inverse

\(X=A^{-1}B\) with \(B=(7,\ 4)\). Then \(x=3(7)-5(4)=1\) and \(y=-1(7)+2(4)=1\).

Check: \(2+5=7\) and \(1+3=4\). \((1,\ 1)\)

Problem 9: Solve a system with the inverse

\(\det A=9-8=1\), so \(A^{-1}\) has rows \(3,\ -4\) and \(-2,\ 3\). Then \(x=3(10)-4(7)=2\) and \(y=-2(10)+3(7)=1\).

\((2,\ 1)\). Check: \(6+4=10\) and \(4+3=7\).

Problem 10: Determinant 11

\(\det A=5(1)-(-2)(3)=11\). \(A^{-1}=\frac{1}{11}\) times rows \(1,\ 2\) and \(-3,\ 5\).

\(x=\frac{1}{11}(1\cdot 4+2\cdot 9)=\frac{22}{11}=2\) and \(y=\frac{1}{11}(-3\cdot 4+5\cdot 9)=\frac{33}{11}=3\).

\((2,\ 3)\). Check: \(10-6=4\) and \(6+3=9\).

Precalculus Matrix Systems Worksheet Answer Key: problems 11–15

Problem 11: A fractional solution

\(\det A=8-6=2\). \(A^{-1}=\frac{1}{2}\) times rows \(2,\ -1\) and \(-6,\ 4\).

\(x=\frac{1}{2}(2\cdot 3-1\cdot 5)=\frac{1}{2}\) and \(y=\frac{1}{2}(-6\cdot 3+4\cdot 5)=1\).

\((\frac{1}{2},\ 1)\). Check: \(2+1=3\) and \(3+2=5\).

Problem 12: When the determinant is zero

Both systems have \(\det A=1(4)-2(2)=0\), so \(A^{-1}\) does not exist.

1. The second equation is twice the first, so the lines coincide: infinitely many solutions.
2. The left sides match but \(2\cdot 3=6\neq 7\), so the lines are parallel: no solution.

(a) infinitely many solutions; (b) no solution

Problem 13: Apply a matrix to a point

1. \((2\cdot 4+0,\ 0+3(-1))=(8,\ -3)\)
2. \((0\cdot 2-1\cdot 5,\ 1\cdot 2+0)=(-5,\ 2)\)
3. \((0-1\cdot 1,\ 1(-3)+0)=(-1,\ -3)\)

(a) \((8,\ -3)\) (b) \((-5,\ 2)\) (c) \((-1,\ -3)\)

Problem 14: Shear the unit square

Columns of \(S\) are the images of \((1,\ 0)\) and \((0,\ 1)\), so \((1,\ 0)\to(1,\ 0)\) and \((0,\ 1)\to(2,\ 1)\). Also \((1,\ 1)\to(1+2,\ 0+1)=(3,\ 1)\).

Images: \((0,\ 0)\), \((1,\ 0)\), \((3,\ 1)\), \((2,\ 1)\). The base and height are both \(1\), so the area is still \(1=\det S\).

Problem 15: Build the matrix from two images

The columns of the matrix are the images: first column \((0,\ 1)\), second column \((-1,\ 0)\). So the matrix has rows \(0,\ -1\) and \(1,\ 0\), a \(90^{\circ}\) counterclockwise rotation.

Image of \((3,\ 2)\): \((0\cdot 3-1\cdot 2,\ 1\cdot 3+0\cdot 2)=(-2,\ 3)\).

Rows \(0,\ -1\) and \(1,\ 0\); image \((-2,\ 3)\)

Precalculus Matrix Systems Worksheet Answer Key: problems 16–20

Problem 16: Reflections

1. \((1\cdot 5+0,\ 0-1\cdot 4)=(5,\ -4)\).
2. The \(x\)-coordinate stays and the \(y\)-coordinate changes sign: a reflection across the \(x\)-axis.
3. \(\det T=1(-1)-0=-1\). The area is unchanged (\(|-1|=1\)); the negative sign shows the orientation is flipped.

(a) \((5,\ -4)\) (b) reflection across the \(x\)-axis (c) \(-1\)

Problem 17: Area scale factor

(a) \(\det A=12-2=10\), so areas are multiplied by \(|10|=10\). The picture confirms it: the image of the unit square has area \(10\).

(b) \(10\times 6=60\). (c) \(35: 10=3.5\).

(a) 10 (b) 60 (c) 3.5

Problem 18: Negative determinant

\(\det B=1(1)-2(3)=-5\). The area scale factor is \(|-5|=5\), so the image has area \(5\times 4=20\). The negative sign means the triangle is flipped (reversed orientation).

Area \(20\); the image is a mirror-like flip.

Problem 19: School store prices

Let \(p\) be the pen price and \(n\) the notebook price. Then \(3p+2n=7.25\) and \(2p+5n=11.25\). \(\det A=15-4=11\); \(A^{-1}=\frac{1}{11}\) times rows \(5,\ -2\) and \(-2,\ 3\).

\(p=\frac{1}{11}(5\cdot 7.25-2\cdot 11.25)=\frac{13.75}{11}=1.25\) and \(n=\frac{1}{11}(-2\cdot 7.25+3\cdot 11.25)=\frac{19.25}{11}=1.75\).

Check: \(3.75+3.50=7.25\) and \(2.50+8.75=11.25\). Pen \(\$1.25\), notebook \(\$1.75\)

Problem 20: Farmers market

With \(a\) and \(b\) the prices: \(4a+3b=19\), \(3a+2b=13.5\). \(\det A=8-9=-1\), so \(A^{-1}\) has rows \(-2,\ 3\) and \(3,\ -4\).

\(a=-2(19)+3(13.5)=-38+40.5=2.5\) and \(b=3(19)-4(13.5)=57-54=3\).

Check: \(10+9=19\) and \(7.5+6=13.5\). Apples \(\$2.50\) per pound, pears \(\$3.00\) per pound

Precalculus Matrix Systems Worksheet Answer Key: problems 21–24

Problem 21: Bake sale with three items

With prices \(c\), \(b\), \(m\): \(A\) has rows \(4,\ 3,\ 2\), \(2,\ 5,\ 1\) and \(1,\ 2,\ 6\); \(B=(17,\ 15.5,\ 20.5)\).

Technology gives \(\det A=77\neq 0\), so \(A^{-1}\) exists, and \(X=A^{-1}B=(1.5,\ 2,\ 2.5)\).

Check day 1: \(6+6+5=17\). Day 2: \(3+10+2.5=15.5\). Day 3: \(1.5+4+15=20.5\). Cookie \(\$1.50\), brownie \(\$2.00\), muffin \(\$2.50\)

Problem 22: Scaling a game sprite

The matrix multiplies \(x\) by 3 and \(y\) by 2: the corner \((5,\ 4)\) goes to \((15,\ 8)\). New size: \(15\) by \(8\), area \(120\).

Old area \(20\); \(\det=6-0=6\); \(6\times 20=120\). New size \(15\) by \(8\), area \(120\)

Problem 23: Mapping a triangular lot

Images: \((4,\ 0)\to(8,\ -4)\); \((0,\ 3)\to(0+3,\ 0+9)=(3,\ 9)\); \((0,\ 0)\to(0,\ 0)\).

Way 1: \(\det=6+1=7\), original area \(\frac{1}{2}(4)(3)=6\), image area \(7\times 6=42\).

Way 2 (shoelace): \(\frac{1}{2}|8\cdot 9-(-4)\cdot 3|=\frac{1}{2}(84)=42\).

Vertices \((0,\ 0)\), \((8,\ -4)\), \((3,\ 9)\); area \(42\)

Problem 24: Undo a transformation

\(\det H=2-1=1\), so \(H^{-1}\) has rows \(1,\ -1\) and \(-1,\ 2\). Then \(P=H^{-1}(11,\ 8)=(11-8,\ -11+16)=(3,\ 5)\).

Check: \(H(3,\ 5)=(6+5,\ 3+5)=(11,\ 8)\). Since \(|\det H|=1\), areas keep their size. \(P=(3,\ 5)\)

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