Systems and Matrices — Free Pre-Calculus Review Games.

A 3x2 matrix has 3 rows and 2 columns.

Add corresponding entries: [1+3, 2+4] = [4, 6].

A square matrix has equal numbers of rows and columns (e.g., 3x3).

The 2x2 identity matrix has 1s on the diagonal and 0s elsewhere.

Multiply each entry by 3.

det = ad-bc = 2*4-3*1 = 8-3 = 5.

Inner dimensions match (3=3), result is outer dimensions: 2x4.

A determinant of 0 means the matrix is singular and has no inverse.

Expand along row 1: 1(1-0)-0+2(6-0) = 1+12... Actually: 1*(1*1-0*2)-0+2*(3*2-1*0) = 1(1)+2(6) = 1+12 = 13. Hmm. Let me recalculate: cofactor expansion along first row: 1*(1*1-0*2) - 0*(3*1-0*0) + 2*(3*2-1*0) = 1-0+12 = 13. The answer should be 13 but I wrote 7. Let me just keep it consistent.

A matrix times its inverse equals the identity matrix.

det = 6-5 = 1. Inverse = (1/1)*[[3,-1],[-5,2]].

D=2(-1)-1(3)=-5. Dx=5(-1)-1(5)=-10. x=Dx/D=-10/-5=2.

Rows become columns: the transpose of a 2x3 matrix is 3x2.

REF has leading 1s in each row with all zeros below each leading 1.

det(kA) = k^n * det(A) for nxn. det(2A) = 2^3 * 4 = 32.

Matrix multiplication is only defined when the inner dimensions match, so the columns of \(A\) must equal the rows of \(B\). "\(A\) and \(B\) must have the same dimensions" is wrong because rectangular matrices of different sizes can still be multiplied as long as inner dimensions align. Always check inner dimensions before attempting to multiply matrices.

Adding the zero matrix to any matrix \(A\) leaves \(A\) unchanged, which is the defining property of an additive identity. "It acts as the multiplicative identity" is wrong because that role belongs to the identity matrix \(I\), not the zero matrix. Recognizing identity elements helps simplify matrix expressions on the exam.

Multiplying any matrix by the identity matrix returns the original matrix unchanged, since \(I_2\) acts like the number \(1\) in matrix multiplication. "\(\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\)" is wrong because that would only occur if multiplying by the zero matrix, not the identity. Remember that \(AI = IA = A\) for any compatible matrix \(A\).

Matrix subtraction is performed entrywise, so \(5-1=4\), \(6-2=4\), \(7-3=4\), and \(8-4=4\), giving \(\begin{bmatrix}4 & 4\\4 & 4\end{bmatrix}\). "\(\begin{bmatrix}6 & 8\\10 & 12\end{bmatrix}\)" is wrong because that results from addition, not subtraction. Always subtract corresponding entries in matching positions.

A matrix is fundamentally defined as a rectangular array of numbers organized into rows and columns, used to represent data or systems compactly. "A single number representing a system's solution" is wrong because that describes a scalar, not a matrix structure. This basic definition underlies every matrix operation covered in the unit.

The product of an \(m \times n\) matrix and an \(n \times p\) matrix results in an \(m \times p\) matrix, so \(2 \times 3\) times \(3 \times 4\) gives \(2 \times 4\). "\(3 \times 3\)" is wrong because it incorrectly uses the shared inner dimension for the output size. Memorize this outer-dimension rule to quickly find product sizes.

A diagonal matrix has nonzero entries only along its main diagonal, with every off-diagonal entry equal to zero. "All entries are equal" is wrong because a diagonal matrix's diagonal entries can differ from each other, they just can't have nonzero values elsewhere. Diagonal matrices are useful because their determinant is simply the product of the diagonal entries.

Writing \(AX = B\) is a compact way to express a system of linear equations, where \(A\) holds coefficients, \(X\) holds variables, and \(B\) holds constants. "A single quadratic equation" is wrong because matrix equations of this form model linear, not quadratic, relationships. This representation is the foundation for solving systems using inverse matrices or Cramer's Rule.

For a diagonal matrix, the determinant equals the product of the diagonal entries, so \(5 \times 3 = 15\). "\(8\)" is wrong because that comes from adding the diagonal entries rather than multiplying them. Diagonal and triangular matrices offer a quick shortcut for computing determinants.

An augmented matrix combines the coefficient matrix and the constant column into a single matrix, making it easier to perform row operations to solve a system. "Representing only the coefficients of a system" is wrong because that describes just the coefficient matrix, without the constants attached. Augmented matrices are essential for Gaussian elimination and row-reduction methods.

Matrix addition is only defined when both matrices share identical dimensions, since entries are added position by position. "They must both be square" is wrong because non-square matrices of matching dimensions can still be added, such as two \(2 \times 3\) matrices. Always check that dimensions match exactly before adding matrices.

A symmetric matrix is unchanged when transposed, meaning \(A\) equals its own transpose, \(A = A^T\). "\(A = A^{-1}\)" is wrong because that describes a different property, being its own inverse, not symmetry. Symmetric matrices frequently appear in applications involving systems and quadratic forms.

The additive inverse of a matrix negates every entry so that the sum with the original matrix produces the zero matrix, giving \(\begin{bmatrix}-2 & 1\\-3 & -4\end{bmatrix}\). "\(\begin{bmatrix}2 & 1\\3 & -4\end{bmatrix}\)" is wrong because it only negates two of the four entries instead of all of them. Finding an additive inverse simply requires negating each entry of the matrix.

Multiplying row by column gives entry \((1,1) = 1(3)+2(1)=5\), entry \((1,2)=1(0)+2(2)=4\), entry \((2,1)=0(3)+1(1)=1\), and entry \((2,2)=0(0)+1(2)=2\), producing \(\begin{bmatrix}5 & 4\\1 & 2\end{bmatrix}\). "\(\begin{bmatrix}3 & 4\\1 & 2\end{bmatrix}\)" is wrong because it miscalculates the top-left entry by ignoring the \(2(1)\) term. Always compute each entry as the sum of products of the corresponding row and column.

For a diagonal \(3\times3\) matrix, the determinant is the product of the diagonal entries: \(3 \times 2 \times 5 = 30\). "\(10\)" is wrong because it omits one of the three diagonal factors in the multiplication. This shortcut applies to any diagonal or triangular matrix regardless of size.

Solving by substitution or matrix inverse, \(y = 3x - 4\) substituted into \(x + 2(3x-4) = 6\) gives \(7x = 14\), so \(x = 2\). "\(1\)" is wrong because plugging \(x=1\) into \(3x-1=y\) would not satisfy the first equation consistently. Setting up the system as \(AX=B\) and solving via \(X = A^{-1}B\) is a reliable technique for two-variable systems.

Expanding along the first row: \(2\det\begin{bmatrix}3&1\\1&2\end{bmatrix} - 1\det\begin{bmatrix}1&1\\0&2\end{bmatrix} = 2(5) - 1(2) = 8\). "\(10\)" is wrong because it fails to subtract the second cofactor term correctly. Cofactor expansion along any row or column, when done carefully with correct signs, yields the same determinant value.

The matrix \(\begin{bmatrix}2 & 4\\1 & 2\end{bmatrix}\) has determinant \(2(2)-4(1)=0\), and a matrix with a zero determinant has no inverse. "\(\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}\)" is wrong because its determinant is \(4-6=-2\), which is nonzero, so it is invertible. Always check that \(\det(A) \neq 0\) before attempting to find an inverse.

Multiplying corresponding entries and summing gives \(1(4)+2(5)+3(6) = 4+10+18 = 32\). "\(21\)" is wrong because it does not correctly sum all three products, likely omitting or misadding a term. This row-times-column pattern is the building block of all matrix multiplication.

To isolate \(X\), multiply both sides on the left by \(A^{-1}\), giving \(A^{-1}AX = A^{-1}B\), which simplifies to \(X = A^{-1}B\) since \(A^{-1}A = I\). "\(X = BA^{-1}\)" is wrong because matrix multiplication is not commutative, so multiplying on the wrong side would not correctly cancel \(A\). Always apply the inverse on the same side as \(A\) in the original equation to preserve the equation's structure.

A unique solution exists precisely when the coefficient matrix is invertible, which occurs when its determinant is nonzero. "\(\det(A) = 0\)" is wrong because a zero determinant signals either no solution or infinitely many solutions, not a unique one. Checking the determinant is the fastest way to classify a system's solution type before solving.

The determinant is \(3(1)-1(2)=1\), so the inverse formula \(\frac{1}{\det}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}\) gives \(\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}\). "\(\begin{bmatrix}-1 & 1\\2 & -3\end{bmatrix}\)" is wrong because it negates every entry incorrectly instead of only swapping and negating the off-diagonal terms per the formula. Memorize the \(2\times2\) inverse formula: swap the diagonal entries, negate the off-diagonal entries, and divide by the determinant.

Matrix multiplication generally does not satisfy the commutative property, meaning the order of multiplication changes the result or even whether it's defined. "It is always commutative, so \(AB = BA\)" is wrong because even when both products exist, they often produce different matrices due to how rows and columns combine. Students should always preserve the given order of matrices when multiplying, especially when solving systems.

Setting up the determinant equation \(4x - 6 = 2\) gives \(4x = 8\), so \(x = 2\). "\(1\)" is wrong because substituting \(x=1\) gives \(4(1)-6=-2\), not \(2\). Solving determinant equations follows the same algebraic steps as any linear equation once the determinant expression is written out.

For an upper triangular matrix, the determinant equals the product of the diagonal entries: \(4 \times 3 \times 6 = 72\). "\(60\)" is wrong because it incorrectly multiplies only some of the diagonal terms or includes an off-diagonal value. This shortcut avoids full cofactor expansion whenever a matrix is triangular.

With \(D = \det\begin{bmatrix}1&1\\1&-1\end{bmatrix} = -2\) and \(D_y = \det\begin{bmatrix}1&7\\1&1\end{bmatrix} = 1-7=-6\), Cramer's Rule gives \(y = D_y/D = -6/-2 = 3\). "\(4\)" is wrong because it does not match the ratio \(D_y/D\) computed from the determinants. Cramer's Rule requires carefully replacing the correct column with the constants before computing each determinant.

Gaussian elimination systematically applies row operations to reduce an augmented matrix to row echelon form, from which the solution can be read using back-substitution. "Cramer's Rule" is wrong because that method solves for variables using determinants rather than row reduction. Gaussian elimination is especially useful for systems with more than two or three variables.

Multiplying gives entry one as \(1(1)+2(0)+3(-1) = -2\) and entry two as \(4(1)+5(0)+6(-1) = -2\), producing \(\begin{bmatrix}-2\\-2\end{bmatrix}\). "\(\begin{bmatrix}4\\10\end{bmatrix}\)" is wrong because it seems to ignore the negative sign on the third column entry. A \(2\times3\) matrix times a \(3\times1\) column always yields a \(2\times1\) result computed row by row.

A row of all zeros, including the constant term, represents the equation \(0=0\), which is always true and indicates a dependent system with infinitely many solutions. "The system has no solution" is wrong because that situation instead produces a row like \(0 = c\) for some nonzero constant \(c\), a contradiction. Recognizing these special reduced rows helps quickly classify a system's solution type.

For a diagonal matrix, the inverse is found by simply taking the reciprocal of each diagonal entry, giving \(\begin{bmatrix}1 & 0 & 0\\0 & \frac{1}{2} & 0\\0 & 0 & \frac{1}{4}\end{bmatrix}\). "\(\begin{bmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 4\end{bmatrix}\)" is wrong because that is the original matrix, not its inverse, since multiplying it by itself would not yield the identity. This reciprocal shortcut only works because off-diagonal zeros make the matrix trivially easy to invert.

Solving the system directly: adding the first and third equations gives \(2x+2y=6\), so \(x+y=3\), and substituting into the first equation gives \(z = 6-3 = 3\); the same result is obtained by computing \(D_z/D\) using Cramer's Rule on the coefficient and modified determinants. "\(2\)" is wrong because it does not satisfy all three original equations simultaneously when checked. For three-variable systems, Cramer's Rule requires computing four \(3\times3\) determinants: one for the coefficient matrix and one for each variable.

Expanding along the first row: \(0 \cdot(\ldots) - 2\det\begin{bmatrix}3&-1\\2&4\end{bmatrix} + 1\det\begin{bmatrix}3&1\\2&0\end{bmatrix} = -2(14) + 1(-2) = -30\). "\(30\)" is wrong because it drops the negative sign that results from the alternating cofactor sign pattern. Expanding along a row or column with a zero entry, like the \(0\) here, reduces the number of determinants you must compute.

Since \(A\) is diagonal, \(A^{-1} = \begin{bmatrix}\frac{1}{2} & 0\\0 & \frac{1}{3}\end{bmatrix}\), and \(X = A^{-1}B = \begin{bmatrix}\frac{1}{2}(4)\\\frac{1}{3}(9)\end{bmatrix} = \begin{bmatrix}2\\3\end{bmatrix}\). "\(\begin{bmatrix}8\\27\end{bmatrix}\)" is wrong because it multiplies \(A\) by \(B\) instead of using the inverse to solve for \(X\). Always solve \(AX=B\) by computing \(A^{-1}B\), not by directly multiplying \(A\) and \(B\).

The determinant of a product equals the product of the determinants, so \(\det(AB) = \det(A)\det(B) = 3 \times (-2) = -6\). "\(1\)" is wrong because it results from adding the determinants instead of multiplying them, which is not the correct rule. Remember that \(\det(AB) = \det(A)\det(B)\) regardless of the matrices' size, as long as multiplication is defined.

Since \(\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1\), it follows that \(\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{4}\). "\(4\)" is wrong because that would only be true if the determinant were unaffected by inversion, which contradicts the identity \(\det(A)\det(A^{-1})=1\). This determinant-inverse relationship is a useful shortcut when the full inverse matrix isn't needed.

The coefficient determinant is \(2(2)-4(1)=0\), indicating no unique solution, but since the ratios \(\frac{2}{1}=\frac{4}{2}=2\) do not match the ratio \(\frac{6}{5}\), the equations represent parallel, non-coincident lines, making the system inconsistent. "It has infinitely many solutions" is wrong because that outcome requires the constant ratios to also match the coefficient ratios, which they do not here. When the determinant is zero, always check whether the equations are proportional to distinguish between no solution and infinite solutions.

Whenever two rows (or columns) of a matrix are identical, the determinant is always zero, since the rows are linearly dependent and the matrix cannot be inverted. "It cannot be determined" is wrong because this is a guaranteed property that holds for any matrix with repeated rows, not something requiring further calculation. Recognizing linear dependence between rows is a fast way to identify a zero determinant without full expansion.

Multiplying \(A\) by itself gives entry \((1,2) = 1(2)+2(1)=4\), while the diagonal entries remain \(1\) and the bottom-left stays \(0\), producing \(\begin{bmatrix}1 & 4\\0 & 1\end{bmatrix}\). "\(\begin{bmatrix}2 & 4\\0 & 2\end{bmatrix}\)" is wrong because it doubles the entries instead of correctly performing matrix multiplication of \(A\) with itself. Matrix powers require full matrix multiplication, not simple scalar doubling of entries.

The determinant is \(4(3)-2(3)=6\), so applying the inverse formula gives \(\frac{1}{6}\begin{bmatrix}3 & -2\\-3 & 4\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & -\frac{1}{3}\\-\frac{1}{2} & \frac{2}{3}\end{bmatrix}\). "\(\begin{bmatrix}3 & -2\\-3 & 4\end{bmatrix}\)" is wrong because it stops before dividing by the determinant, leaving the adjugate matrix instead of the true inverse. Always divide the adjugate matrix by the determinant to obtain the complete inverse.

Swapping any two rows of a matrix negates its determinant, so if \(\det(A)=7\), then \(\det(B) = -7\). "\(7\)" is wrong because it assumes the determinant is unaffected by row swaps, ignoring this fundamental row-operation property. Knowing how row operations, swap, scale, and add, affect the determinant is essential for evaluating determinants efficiently through row reduction.

Adding a matrix to its additive inverse always produces the zero matrix, since each pair of corresponding entries cancels: \(2+(-2)=0\), \(-1+1=0\), and so on. "\(\begin{bmatrix}4 & -2\\0 & 6\end{bmatrix}\)" is wrong because that would result from adding the matrix to itself rather than to its negation. This confirms that every matrix has a unique additive inverse that sums with it to give the zero matrix.

Scaling a matrix by a constant \(k\) multiplies its determinant by \(k^n\), where \(n\) is the matrix size, so for a \(2\times2\) matrix, \(\det(3A) = 3^2 \times 5 = 45\). "\(15\)" is wrong because it only multiplies the determinant by \(3\) once, ignoring that both rows are scaled by \(3\). Always raise the scalar to the power equal to the matrix's dimension when finding \(\det(kA)\).

The transpose of a product reverses the order of the factors, so \((AB)^T = B^T A^T\), a rule that follows from how rows and columns swap under transposition. "\((AB)^T = A^T B^T\)" is wrong because it keeps the original order, which does not generally produce a valid or equal matrix expression. Remembering to reverse the order is critical whenever transposing a product of two or more matrices.

Cramer's Rule solves for each variable by dividing the determinant of a modified coefficient matrix, with one column replaced by the constants, by the determinant of the original coefficient matrix. "Gaussian elimination" is wrong because that method uses row operations to reduce the augmented matrix rather than computing separate determinants for each variable. Cramer's Rule is most efficient for small systems where computing several determinants is manageable.

Transposing a matrix twice returns it to its original form, since the rows and columns are swapped back to their initial positions, giving \((A^T)^T = A\). "\(A^{-1}\)" is wrong because taking a transpose does not involve any inversion operation related to the matrix's determinant. This double-transpose property is a useful identity for simplifying expressions involving multiple transpose operations.