Algebra 3 Quiz: Matrix Solutions To Systems
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Matrix Solutions To SystemsQuestion 1 of 20

The system below is written as the matrix equation AX=BAX = B, where A=[12−1213312]A=\begin{bmatrix}1&2&-1\\2&1&3\\3&1&2\end{bmatrix} and B=[525]B=\begin{bmatrix}5\\2\\5\end{bmatrix}. If A−1=110[−1−5755−5−15−3]A^{-1}=\frac{1}{10}\begin{bmatrix}-1&-5&7\\5&5&-5\\-1&5&-3\end{bmatrix}, what is the zz-coordinate of the solution X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}?

2
1
-1
-10
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Algebra 3 Quiz

Algebra 3 Quiz: Matrix Solutions To Systems

Practice Matrix Solutions To Systems in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Matrix Solutions To Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The system below is written as the matrix equation AX=BAX = B, where A=[12−1213312]A=\begin{bmatrix}1&2&-1\\2&1&3\\3&1&2\end{bmatrix} and B=[525]B=\begin{bmatrix}5\\2\\5\end{bmatrix}. If A−1=110[−1−5755−5−15−3]A^{-1}=\frac{1}{10}\begin{bmatrix}-1&-5&7\\5&5&-5\\-1&5&-3\end{bmatrix}, what is the zz-coordinate of the solution X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}?

  1. 2
  2. 1
  3. -1 (correct answer)
  4. -10
Explanation: When you see a matrix equation AX=BAX=B and you already have A−1A^{-1}, your move is to multiply both sides on the left by A−1A^{-1}. That gives X=A−1BX=A^{-1}B, so the solution vector is just that product. Since the question asks for the zz-coordinate, you only need the third row of A−1A^{-1}. Compute the third row of A−1A^{-1} dotted with BB: z=110[−15−3][525]=110(−5+10−15)=−1010=−1.z = \frac{1}{10}\begin{bmatrix}-1 & 5 & -3\end{bmatrix} \begin{bmatrix}5\\2\\5\end{bmatrix} = \frac{1}{10}(-5+10-15) = \frac{-10}{10} = -1. So the zz-coordinate is −1-1. The other choices come from common missteps. The value 22 is actually the xx-coordinate: the first row gives x=2x=2. The value 11 is the yy-coordinate, from the second row. Both are real parts of the solution, but neither is zz. The value −10-10 is the numerator inside the calculation before dividing by 1010; if you forget the 110\frac{1}{10} factor, you'd incorrectly choose −10-10. A useful habit: when XX has multiple components, identify exactly which row of A−1A^{-1} corresponds to the variable you need. Then compute only that row times BB. This saves time and reduces arithmetic errors. Also, always check whether the final step includes dividing by the determinant factor—here the 110\frac{1}{10} is easy to drop.

Question 2

Solve x+2y=5x+2y=5 and 3x+5y=133x+5y=13 by finding A−1A^{-1} and computing X=A−1BX=A^{-1}B.

  1. (−1,−2)(-1,-2)
  2. (1,2)(1,2) (correct answer)
  3. (2,1)(2,1)
  4. (9,−2)(9,-2)
Explanation: Here det⁡(A)=1(5)−2(3)=−1\det(A)=1(5)-2(3)=-1, so A−1=[−523−1]A^{-1}=\begin{bmatrix}-5&2\\3&-1\end{bmatrix} and X=A−1B=(1,2)X=A^{-1}B=(1,2), which checks in both equations. The pair (−1,−2)(-1,-2) comes from using the adjugate without dividing by the negative determinant. The pair (2,1)(2,1) reverses the coordinates. The pair (9,−2)(9,-2) comes from a sign slip in elimination that makes y=−2y=-2.

Question 3

A student tries to solve AX=BAX=B with [2−6−39]\begin{bmatrix}2&-6\\-3&9\end{bmatrix} by computing A−1A^{-1}. Why does this approach fail?

  1. The matrix AA is not square, so no inverse is defined.
  2. The constant matrix BB contains a negative entry.
  3. The system has more variables than it has equations.
  4. The determinant 2(9)−(−6)(−3)=02(9)-(-6)(-3)=0, so AA has no inverse. (correct answer)
Explanation: A square matrix has an inverse only when its determinant is nonzero, and here the determinant is 18−18=018-18=0. The claim that AA is not square is false, since AA is 2×22\times 2. The entries of BB never affect whether A−1A^{-1} exists. The system has two equations in two unknowns, so the count of variables and equations matches.

Question 4

Back-substitute to solve the system whose augmented matrix in row echelon form is [13501−2]\left[\begin{array}{cc|c}1&3&5\\0&1&-2\end{array}\right].

  1. (−1,−2)(-1,-2)
  2. (5,−2)(5,-2)
  3. (11,−2)(11,-2) (correct answer)
  4. (11,2)(11,2)
Explanation: The second row gives y=−2y=-2, and substituting into x+3y=5x+3y=5 gives x=5+6=11x=5+6=11, so the solution is (11,−2)(11,-2). The pair (5,−2)(5,-2) skips the back-substitution and reads xx straight from the constant column. The pair (−1,−2)(-1,-2) subtracts 66 instead of adding it. The pair (11,2)(11,2) drops the negative sign on yy.

Question 5

A system in xx, yy, and zz has the row echelon augmented matrix [1−21001−3−10012]\left[\begin{array}{ccc|c}1&-2&1&0\\0&1&-3&-1\\0&0&1&2\end{array}\right]. What is its solution?

  1. (8,5,2)(8,5,2) (correct answer)
  2. (0,−1,2)(0,-1,2)
  3. (−12,5,2)(-12,5,2)
  4. (2,5,8)(2,5,8)
Explanation: The bottom row gives z=2z=2; then y−3(2)=−1y-3(2)=-1 gives y=5y=5; then x−2(5)+2=0x-2(5)+2=0 gives x=8x=8. The triple (0,−1,2)(0,-1,2) reads the constant column directly without back-substituting. The triple (−12,5,2)(-12,5,2) mishandles the sign when isolating xx. The triple (2,5,8)(2,5,8) lists the correct values in reverse order.

Question 6

Given A−1=[2−1−32]A^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix} and B=[45]B=\begin{bmatrix}4\\5\end{bmatrix} , compute X=A−1BX=A^{-1}B.

  1. (−3,2)(-3,2)
  2. (2,3)(2,3)
  3. (3,−2)(3,-2) (correct answer)
  4. (13,22)(13,22)
Explanation: Row-by-column multiplication gives 2(4)+(−1)(5)=32(4)+(-1)(5)=3 and (−3)(4)+2(5)=−2(-3)(4)+2(5)=-2, so X=(3,−2)X=(3,-2). The pair (13,22)(13,22) comes from ignoring the negative signs and adding every product. The pair (−3,2)(-3,2) reads the first column of A−1A^{-1} rather than computing the product. The pair (2,3)(2,3) reverses the two entries and drops the sign.

Question 7

Evaluate the determinant of $$ \begin{bmatrix}2&1&0\1&3&1\0&1&2\end{bmatrix}

  1. 00
  2. 88 (correct answer)
  3. 1010
  4. 1212
Explanation: Expanding along the first row gives 2(3⋅2−1⋅1)−1(1⋅2−1⋅0)+0=10−2=82(3\cdot 2-1\cdot 1)-1(1\cdot 2-1\cdot 0)+0=10-2=8. The value 1010 keeps only the first term and ignores the minor from the entry 11. The value 1212 adds that second term instead of subtracting it, since the cofactor signs alternate. The value 00 assumes the zero entries force a zero determinant, which they do not.

Question 8

Which augmented matrix represents 3x−y=73x-y=7 together with x+4y=−2x+4y=-2?

  1. [3−1714−2]\left[\begin{array}{cc|c}3&-1&7\\1&4&-2\end{array}\right] (correct answer)
  2. [31714−2]\left[\begin{array}{cc|c}3&1&7\\1&4&-2\end{array}\right]
  3. [3−1−2147]\left[\begin{array}{cc|c}3&-1&-2\\1&4&7\end{array}\right]
  4. [−13741−2]\left[\begin{array}{cc|c}-1&3&7\\4&1&-2\end{array}\right]
Explanation: Each row lists that equation's xx coefficient, yy coefficient, and constant in order, so the first row is 3, −1, 73,\,-1,\,7 and the second is 1, 4, −21,\,4,\,-2. The version beginning 3, 13,\,1 loses the negative on −y-y. The version with constants −2-2 and 77 attaches each constant to the wrong equation. The version beginning −1, 3-1,\,3 reverses the coefficient columns.

Question 9

Row reduce to solve x+y+z=6x+y+z=6, 2x−y+z=32x-y+z=3, and x+2y−z=2x+2y-z=2.

  1. (3,2,1)(3,2,1)
  2. (2,1,3)(2,1,3)
  3. (1,3,2)(1,3,2)
  4. (1,2,3)(1,2,3) (correct answer)
Explanation: Eliminating zz leads to x−2y=−3x-2y=-3 and 2x+3y=82x+3y=8, giving y=2y=2, x=1x=1, and z=3z=3; all three original equations check. The triple (3,2,1)(3,2,1) reverses the order of the values. The triple (2,1,3)(2,1,3) swaps xx and yy, failing 2x−y+z=32x-y+z=3. The triple (1,3,2)(1,3,2) swaps yy and zz, failing x+2y−z=2x+2y-z=2.

Question 10

For which value of kk does $$ \begin{bmatrix}3&k\2&4\end{bmatrix}

  1. −6-6
  2. 00
  3. 66 (correct answer)
  4. 1212
Explanation: The inverse fails to exist exactly when the determinant is zero, and 3(4)−k(2)=12−2k=03(4)-k(2)=12-2k=0 gives k=6k=6. The value −6-6 comes from solving 12+2k=012+2k=0 with the wrong sign. The value 1212 comes from setting 2k=122k=12 but forgetting to divide by 22. The value 00 leaves the determinant equal to 1212, which is nonzero.

Question 11

The coefficient matrix of a system is the diagonal matrix $$ \begin{bmatrix}1&0&0\0&2&0\0&0&5\end{bmatrix}

  1. [10001200015]\begin{bmatrix}1&0&0\\0&\frac{1}{2}&0\\0&0&\frac{1}{5}\end{bmatrix} (correct answer)
  2. [−1000−2000−5]\begin{bmatrix}-1&0&0\\0&-2&0\\0&0&-5\end{bmatrix}
  3. [100020005]\begin{bmatrix}1&0&0\\0&2&0\\0&0&5\end{bmatrix}
  4. [00150120100]\begin{bmatrix}0&0&\frac{1}{5}\\0&\frac{1}{2}&0\\1&0&0\end{bmatrix}
Explanation: For a diagonal matrix with nonzero entries, the inverse is diagonal with each entry replaced by its reciprocal, and multiplying the two matrices returns the identity. Negating the entries gives a product with −1-1 down the diagonal, not 11. Repeating the original matrix gives a product with 1, 4, 251,\,4,\,25 down the diagonal. Reversing the order of the reciprocals moves nonzero entries off the diagonal, so the product is not the identity.

Question 12

The augmented matrix of a system in xx, yy, and zz reduces to [10−23014−10000]\left[\begin{array}{ccc|c}1&0&-2&3\\0&1&4&-1\\0&0&0&0\end{array}\right]. Describe its solution set.

  1. There is no solution, because the third row is entirely zeros.
  2. There are infinitely many solutions, one for each value of zz. (correct answer)
  3. There is exactly one solution, namely the triple (3,−1,0)(3,-1,0).
  4. There is exactly one solution, namely the triple (3,−1,1)(3,-1,1).
Explanation: Two pivots serve three variables, so zz is free and every choice of zz yields a solution x=3+2zx=3+2z, y=−1−4zy=-1-4z. A row of all zeros signals a dependent equation, not an inconsistency; inconsistency would require a nonzero constant in that row. The triple (3,−1,0)(3,-1,0) is only the particular solution obtained when z=0z=0, and (3,−1,1)(3,-1,1) does not even satisfy the first row, since 3−2(1)=1≠33-2(1)=1\neq 3.

Question 13

Consider the matrix equation [12−124−2111][xyz]=[352]. \begin{bmatrix}1&2&-1\\2&4&-2\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\5\\2\end{bmatrix}. Which conclusion follows?

  1. The system has no solution. (correct answer)
  2. The system has exactly one solution.
  3. The system has infinitely many solutions.
  4. The system has exactly one solution: (1,1,0)(1,1,0).
Explanation: Whenever you have a matrix equation, check whether one row is a multiple of another before doing elimination. Here the first two coefficient rows are proportional: Row 2 is exactly twice Row 1. That means the two equations express the same left-hand side in different ways: Row 1 gives x+2y−z=3x+2y-z=3, and Row 2 gives 2x+4y−2z=52x+4y-2z=5, which simplifies to x+2y−z=52x+2y-z=\tfrac{5}{2}. The same expression cannot equal both 33 and 52\tfrac{5}{2} simultaneously, so no triple (x,y,z)(x,y,z) can satisfy the system. Therefore the conclusion that the system has no solution is forced. The statement that the system has exactly one solution would require a consistent system, but the first two rows are incompatible, so zero solutions exist. The proposed unique solution (1,1,0)(1,1,0) looks tempting because it satisfies the first and third equations: 1+2−0=31+2-0=3 and 1+1+0=21+1+0=2. However, it fails the second equation: 2(1)+4(1)−2(0)=62(1)+4(1)-2(0)=6, not 55. The infinitely many solutions conclusion is another trap: proportional coefficient rows often signal redundancy or free variables, but here the constants are inconsistent, so the system is inconsistent, not underdetermined. Whenever proportional coefficient rows appear, compare the constants. If one row is a multiple of another but the constants are not in the same ratio, the system has no solution. This quick check saves time and prevents a false "infinitely many solutions" conclusion.

Question 14

Which single row operation turns [2461−35]\left[\begin{array}{cc|c}2&4&6\\1&-3&5\end{array}\right] into [1231−35]\left[\begin{array}{cc|c}1&2&3\\1&-3&5\end{array}\right]?

  1. R1→R1−R2R_1 \to R_1 - R_2
  2. R1↔R2R_1 \leftrightarrow R_2
  3. R1→12R1R_1 \to \tfrac{1}{2}R_1 (correct answer)
  4. R1→2R1R_1 \to 2R_1
Explanation: Every entry of the first row is halved, since 2, 4, 62,\,4,\,6 becomes 1, 2, 31,\,2,\,3, while the second row is untouched. The operation R1→R1−R2R_1 \to R_1 - R_2 would produce 1, 7, 11,\,7,\,1 instead. Swapping the rows would put 1, −3, 51,\,-3,\,5 on top. Doubling the first row would give 4, 8, 124,\,8,\,12, moving away from the target rather than toward it.

Question 15

A jeweler's system is 4x+3y=184x+3y=18 and 2x+5y=162x+5y=16. Using the inverse of the coefficient matrix, what is the solution?

  1. (2,3)(2,3)
  2. (3,−2)(3,-2)
  3. (42,28)(42,28)
  4. (3,2)(3,2) (correct answer)
Explanation: With det⁡(A)=4(5)−3(2)=14\det(A)=4(5)-3(2)=14, the inverse is 114[5−3−24]\tfrac{1}{14}\begin{bmatrix}5&-3\\-2&4\end{bmatrix} , and 114(42,28)=(3,2)\tfrac{1}{14}(42,28)=(3,2), which satisfies both equations. The pair (42,28)(42,28) omits the factor 114\tfrac{1}{14}. The pair (2,3)(2,3) reverses the coordinates. The pair (3,−2)(3,-2) carries a sign error into yy and fails the second equation.

Question 16

A system of three equations in two unknowns is written as AX=BAX=B. What are the dimensions of the three matrices?

  1. AA is 3×23\times 2, XX is 2×12\times 1, BB is 3×13\times 1 (correct answer)
  2. AA is 2×32\times 3, XX is 3×13\times 1, BB is 2×12\times 1
  3. AA is 3×33\times 3, XX is 3×13\times 1, BB is 3×13\times 1
  4. AA is 2×22\times 2, XX is 2×12\times 1, BB is 2×12\times 1
Explanation: The coefficient matrix has one row per equation and one column per variable, so it is 3×23\times 2; the variable matrix lists the two unknowns as 2×12\times 1; and the constant matrix lists the three constants as 3×13\times 1. Making AA into 2×32\times 3 reverses the roles of equations and variables. Making AA into 3×33\times 3 invents a third variable. Making AA into 2×22\times 2 discards the third equation entirely.

Question 17

Use the inverse of the coefficient matrix to solve 3x−2y=83x-2y=8 and x+y=1x+y=1.

  1. (−2,1)(-2,1)
  2. (−1,2)(-1,2)
  3. (10,−5)(10,-5)
  4. (2,−1)(2,-1) (correct answer)
Explanation: The determinant is 3(1)−(−2)(1)=53(1)-(-2)(1)=5, so A−1=15[12−13]A^{-1}=\tfrac{1}{5}\begin{bmatrix}1&2\\-1&3\end{bmatrix} and X=15(10,−5)=(2,−1)X=\tfrac{1}{5}(10,-5)=(2,-1), which satisfies both equations. The pair (10,−5)(10,-5) omits the factor 15\tfrac{1}{5}, treating the determinant as 11. The pairs (−2,1)(-2,1) and (−1,2)(-1,2) reverse or negate the coordinates and fail the first equation.

Question 18

In exactly one of the systems below, the coefficient matrix has no inverse. Which system is it?

  1. 2x+3y=52x+3y=5 and 4x−y=14x-y=1
  2. x+3y=4x+3y=4 and 2x+6y=92x+6y=9 (correct answer)
  3. 5x−y=25x-y=2 and x+2y=7x+2y=7
  4. 3x+y=63x+y=6 and x−4y=2x-4y=2
Explanation: An inverse fails to exist only when the determinant is zero, and 1(6)−3(2)=01(6)-3(2)=0 for the system whose second equation is 2x+6y=92x+6y=9, whose left side is exactly twice the first. The others have nonzero determinants: 2(−1)−3(4)=−142(-1)-3(4)=-14, 5(2)−(−1)(1)=115(2)-(-1)(1)=11, and 3(−4)−1(1)=−133(-4)-1(1)=-13, so each of those can be solved by the inverse method.

Question 19

Find the column matrix XX satisfying $$ \begin{bmatrix}4&1\3&-2\end{bmatrix}X=\begin{bmatrix}5\12\end{bmatrix}

  1. [2−3]\begin{bmatrix}2\\-3\end{bmatrix} (correct answer)
  2. [−32]\begin{bmatrix}-3\\2\end{bmatrix}
  3. [3−2]\begin{bmatrix}3\\-2\end{bmatrix}
  4. [22−33]\begin{bmatrix}22\\-33\end{bmatrix}
Explanation: The determinant is 4(−2)−1(3)=−114(-2)-1(3)=-11, and multiplying by the inverse gives X=111(22,−33)X=\tfrac{1}{11}(22,-33), or [2−3]\begin{bmatrix}2\\-3\end{bmatrix}; substituting confirms 4(2)+(−3)=54(2)+(-3)=5 and 3(2)−2(−3)=123(2)-2(-3)=12. The entry pair [22−33]\begin{bmatrix}22\\-33\end{bmatrix} skips dividing by the determinant. The pairs [−32]\begin{bmatrix}-3\\2\end{bmatrix} and $$ \begin{bmatrix}3\-2\end{bmatrix}

Question 20

In the equation AX=BAX=B, the matrix AA is 2×22\times 2 and BB is 2×12\times 1. Why is the expression BA−1BA^{-1} unusable here?

  1. Because A−1A^{-1} never exists when BB is a column matrix.
  2. Because BA−1BA^{-1} equals A−1BA^{-1}B only when AA is symmetric.
  3. Because a 2×12\times 1 matrix times a 2×22\times 2 matrix is an undefined product. (correct answer)
  4. Because BA−1BA^{-1} yields a 2×22\times 2 matrix instead of a column matrix.
Explanation: Multiplication requires the columns of the left factor to match the rows of the right factor, and BB has one column while A−1A^{-1} has two rows, so the product never forms. Whether A−1A^{-1} exists depends only on the determinant of AA, not on the shape of BB. Symmetry of AA does not make the two orders interchangeable. Saying the result is the wrong shape presumes a product that cannot be computed at all.