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Basic Matrices
Understanding Matrices
A matrix is a rectangular arrangement of numbers organized into rows and columns. Each number in the grid is called an element, and its position is identified by its row index and column index.
$$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$$
A matrix A of size m × n has m rows and n columns. Each element a_{ij} sits at row i, column j.
$A$=Name of the matrix(—)
$a_{ij}$=Element in row i, column j(depends on context)
$m \times n$=Dimensions — rows first, then columns(count)
$m = n$
→The matrix is square — equal rows and columns.
Row-Column Convention: Always state dimensions as rows first, then columns. A 3×2 matrix has 3 rows and 2 columns.
Element Addressing: $a_{23}$ means the element in row 2, column 3 — row comes first, column second.
Notation Alternatives: Matrices can also be written with parentheses ( ) instead of square brackets .
Matrices come in several important types, each with special properties that make them useful in different situations.
Row Matrix: A single row of numbers, e.g., $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ — size 1×n.
Column Matrix: A single column of numbers, e.g., $\begin{bmatrix} 4 \\ 5 \end{bmatrix}$ — size m×1.
Square Matrix: Equal rows and columns (n×n). Only square matrixes have determinants and inverse matrixes.
Zero Matrix: Every element is 0, denoted $O$. Adding $O$ to any matrix leaves it unchanged.
Identity Matrix: Square matrix with 1s on the main diagonal and 0s everywhere else, denoted $I$. Multiplying any compatible matrix by $I$ leaves it unchanged — the matrix equivalent of the number 1.
Quick Reference: Identity Matrices
•
$I_1 = \begin{bmatrix} 1 \end{bmatrix}$
•
$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
•
$I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Matrix Addition and Subtraction
Two matrices can be added or subtracted element by element, but only when they have the exact same dimensions.
$$C = A \pm B \implies c_{ij} = a_{ij} \pm b_{ij}$$
Each element in the result is simply the sum or difference of the corresponding elements from the two matrices.
$c_{ij}$=Element at row i, column j of the result matrix C(depends on context)
$a_{ij}, b_{ij}$=Corresponding elements from matrices A and B(depends on context)
Same Size Required: A 2×3 matrix can only be added to another 2×3 matrix. Adding a 2×2 to a 2×3 is undefined.
Element-by-Element: There is no cross-multiplication or interaction between elements — just direct addition or subtraction at matching positions.
Properties: Matrix addition is commutative ($A + B = B + A$) and associative ($(A + B) + C = A + (B + C)$).
Scalar Multiplication
Multiplying a matrix by a scalar means multiplying every single element by that number. The resulting matrix has the same dimensions.
$$kA = \begin{bmatrix} k \cdot a_{11} & k \cdot a_{12} \\ k \cdot a_{21} & k \cdot a_{22} \end{bmatrix}$$
The scalar distributes to every element of the matrix, producing a new matrix of the same size.
$k$=A scalar (real number)(real number)
$A$=Any matrix(—)
Distributes to Every Element: Each $a_{ij}$ becomes $k \cdot a_{ij}$. The matrix size stays the same.
Negative Scalars: Multiplying by $-1$ flips the sign of every element, giving the additive inverse $-A$.
Scalar Factor-Out: $k(A + B) = kA + kB$ — a scalar distributes across matrix addition.
Matrix Multiplication
Matrix multiplication uses the dot product of rows and columns: multiply corresponding elements in a row of the first matrix with a column of the second, then sum them all up.
$$c_{ij} = \sum_{k=1}^{p} a_{ik} \cdot b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{ip}b_{pj}$$
To find the element at row i, column j of the product, take row i from the first matrix and column j from the second, multiply matching pairs, and add all products.
$c_{ij}$=Element at row i, column j of the result(depends on context)
$a_{ik}$=Element in row i, column k of matrix A(depends on context)
$b_{kj}$=Element in row k, column j of matrix B(depends on context)
$AB$ vs. $BA$
→In general $AB \neq BA$ — matrix multiplication is NOT commutative.
Dimension Rule: If A is m×p and B is p×n, then AB is m×n. The inner dimensions must match (columns of A = rows of B).
Step-by-Step Method: For each element of the result, run your finger across a row of A and down a column of B, multiplying pairs and accumulating the total.
Zero Product Trap: $AB = O$ does NOT mean $A = O$ or $B = O$. Two non-zero matrices can multiply to give a zero matrix.
Matrix multiplication has properties that differ from regular number multiplication in important ways.
Not Commutative: $AB \neq BA$ in general. Always multiply in the given order.
Associative: $A(BC) = (AB)C$ — you can regroup multiplications.
Distributive: $A(B + C) = AB + AC$ — multiplication distributes over addition.
Identity Property: $AI = IA = A$ — multiplying by the identity matrix leaves a matrix unchanged.
The Determinant
The determinant is a single number computed from a square matrix. For a 2×2 matrix, it measures the area scaling factor of the transformation the matrix represents.
$$\det(A) = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$
Multiply the main diagonal elements and subtract the product of the off-diagonal elements.
$a, d$=Main diagonal elements (top-left to bottom-right)(depends on context)
$b, c$=Off-diagonal elements(depends on context)
$\det(A)$ or $|A|$=Determinant of matrix A(scalar)
$\det(A) = 0$
→The matrix is singular matrix — it has no inverse and the system may have no unique solution.
$\det(A) \neq 0$
→The matrix is non-singular — it has a unique inverse.
Visual Pattern: Cross-multiply: $ad$ (down-right) minus $bc$ (up-right) — draw an X through the matrix.
Determinant of Identity: $\det(I_n) = 1$ for any size identity matrix.
Determinant of Zero Matrix: $\det(O) = 0$.
For a 3×3 matrix, the determinant is found by cofactor expansion along any row or column. Expanding along the first row is the most straightforward approach.
$$\det(A) = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$$
Expand along the first row: multiply each element by its cofactor (a signed 2×2 minor), then sum the results.
$a_{11}, a_{12}, a_{13}$=Elements of the first row(depends on context)
$\begin{vmatrix} \cdots \end{vmatrix}$=Minor — the 2×2 determinant remaining after removing the element's row and column(scalar)
A row or column is all zeros
→$\det = 0$ automatically.
Two rows are identical
→$\det = 0$.
Sign Pattern: The signs alternate starting with $+$ for position (1,1): $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$.
Minor Method: For each element, cross out its row and column, compute the 2×2 determinant of what remains, then apply the sign from the pattern.
Shortcut (Sarrus' Rule): Rewrite the first two columns to the right of the matrix, then sum the products of three downward diagonals minus three upward diagonals. Works only for 3×3.
Inverse of a 2×2 Matrix
The inverse matrix of A, written $A^{-1}$, is the matrix that satisfies $AA^{-1} = I$. A matrix has an inverse only if its determinant is non-zero.
$$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
Swap the main diagonal elements, negate the off-diagonal elements, then divide every element by the determinant.
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$=The original 2×2 matrix(—)
$\det(A) = ad - bc$=Determinant — must be non-zero(scalar)
$A^{-1}$=Inverse of A(—)
Swap and Negate: Swap $a$ and $d$ on the diagonal, then change the signs of $b$ and $c$. This creates the adjugate matrix.
Division by Determinant: Every element of the adjugate gets divided by $\det(A)$.
Verification Check: Multiply $AA^{-1}$ — you should get $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Non-Invertible Case: If $\det(A) = 0$, the inverse does not exist because of division by zero.
Solving Systems with Matrices
A system of linear equations can be written compactly in matrix form. For a 2×2 system, the solution uses the inverse matrix of the coefficient matrix.
$$A\vec{x} = \vec{b} \implies \vec{x} = A^{-1}\vec{b}$$
Write the coefficients as matrix A, the variables as a column vector, and the constants as another column vector. The solution is found by multiplying the inverse of A with the constant vector.
$A$=Coefficient matrix (from the variables' coefficients)(—)
$\vec{x}$=Column vector of unknowns(—)
$\vec{b}$=Column vector of constants (right-hand side)(—)
Setting Up the Matrix: For the system $ax + by = e$ and $cx + dy = f$, the coefficient matrix is $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, with $\vec{x} = \begin{bmatrix} x \\ y \end{bmatrix}$ and $\vec{b} = \begin{bmatrix} e \\ f \end{bmatrix}$.
Unique Solution Condition: A unique solution exists if and only if $\det(A) \neq 0$.
No Solution or Infinite Solutions: If $\det(A) = 0$, the system either has no solution or infinitely many — there is no unique answer.
Application Strategy
Worked Examples
Problem Statement
A store tracks daily sales of three products across two locations. Location 1 sold 2, 5, 3 units and Location 2 sold 1, 4, 6 units on a given day. What is the total sales matrix across both locations?
matrix addition
Strategic Toolkit
$$C = A + B$$
Step 1
Write the matrices
Represent each location's sales as a 1×3 row matrix.
$A = \begin{bmatrix} 2 & 5 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 4 & 6 \end{bmatrix}$
Step 2
Add corresponding elements
Add element by element since both matrices are 1×3.
$C = \begin{bmatrix} 2+1 & 5+4 & 3+6 \end{bmatrix} = \begin{bmatrix} 3 & 9 & 9 \end{bmatrix}$
Problem Statement
A company's quarterly expenses for three categories are 12, 8, 15 thousand rupees. If expenses increase by 20% next quarter, what are the projected expenses?
scalar multiplication
Strategic Toolkit
$$kA = 1.2A$$
Step 1
Identify the scalar
A 20% increase means multiplying by 1.2.
$k = 1 + 0.20 = 1.2$
Step 2
Multiply each element by the scalar
Multiply every element of the expense matrix by 1.2.
$1.2 \begin{bmatrix} 12 & 8 & 15 \end{bmatrix} = \begin{bmatrix} 14.4 & 9.6 & 18 \end{bmatrix}$
Problem Statement
Find the product $AB$ where $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}$.
matrix multiplication
Strategic Toolkit
$$c_{ij} = \sum_{k} a_{ik} b_{kj}$$
Step 1
Check dimensions
A is 2×2 and B is 2×2, so the product AB will be 2×2. Inner dimensions match.
$(2 \times 2)(2 \times 2) \rightarrow 2 \times 2$
Step 2
Compute row 1, column 1
Dot product of row 1 of A with column 1 of B.
$c_{11} = (2)(5) + (3)(2) = 10 + 6 = 16$
Step 3
Compute row 1, column 2
Dot product of row 1 of A with column 2 of B.
$c_{12} = (2)(1) + (3)(3) = 2 + 9 = 11$
Step 4
Compute row 2, column 1
Dot product of row 2 of A with column 1 of B.
$c_{21} = (1)(5) + (4)(2) = 5 + 8 = 13$
Step 5
Compute row 2, column 2
Dot product of row 2 of A with column 2 of B.
$c_{22} = (1)(1) + (4)(3) = 1 + 12 = 13$
Step 6
Assemble the result
Combine all four elements into the product matrix.
$AB = \begin{bmatrix} 16 & 11 \\ 13 & 13 \end{bmatrix}$
Problem Statement
A bakery makes 3 products from 2 ingredients. The recipe matrix is $R = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 2 & 1 \end{bmatrix}$ (units of each ingredient per product) and the order column vector is $Q = \begin{bmatrix} 10 \\ 5 \\ 8 \end{bmatrix}$ (units ordered per product). Find the total ingredient quantities needed.
matrix multiplicationdimension compatibility
Strategic Toolkit
$$\text{Ingredients} = RQ$$
Step 1
Check dimensions
R is 2×3 and Q is 3×1, so the product is 2×1 — a column vector of ingredient totals.
$(2 \times 3)(3 \times 1) \rightarrow 2 \times 1$
Step 2
Compute ingredient 1 total
Dot product of row 1 of R with column 1 of Q.
$(2)(10) + (1)(5) + (3)(8) = 20 + 5 + 24 = 49$
Step 3
Compute ingredient 2 total
Dot product of row 2 of R with column 1 of Q.
$(1)(10) + (2)(5) + (1)(8) = 10 + 10 + 8 = 28$
Step 4
State the result
The bakery needs 49 units of ingredient 1 and 28 units of ingredient 2.
$RQ = \begin{bmatrix} 49 \\ 28 \end{bmatrix}$
Problem Statement
Find the determinant of $A = \begin{bmatrix} 6 & 4 \\ 3 & 2 \end{bmatrix}$ and state whether A is invertible.
determinant
Strategic Toolkit
$$\det(A) = ad - bc$$
Step 1
Identify the elements
Label a, b, c, d from the matrix positions.
$a = 6, \; b = 4, \; c = 3, \; d = 2$
Step 2
Apply the formula
Multiply the main diagonal and subtract the off-diagonal product.
$\det(A) = (6)(2) - (4)(3) = 12 - 12 = 0$
Step 3
Interpret the result
Since the determinant is zero, the matrix is singular and has no inverse.
$\det(A) = 0 \implies A \text{ is not invertible}$
Problem Statement
Find the determinant of $A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 1 \\ 5 & 2 & 1 \end{bmatrix}$ using cofactor expansion along the first row.
determinant
Strategic Toolkit
$$\det(A) = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}$$
Step 1
Expand along row 1
Write the cofactor expansion with alternating signs: +, −, +.
$\det(A) = 2\begin{vmatrix} 4 & 1 \\ 2 & 1 \end{vmatrix} - 1\begin{vmatrix} 0 & 1 \\ 5 & 1 \end{vmatrix} + 3\begin{vmatrix} 0 & 4 \\ 5 & 2 \end{vmatrix}$
Step 2
Compute each 2×2 minor
Evaluate each of the three 2×2 determinants.
$M_{11} = (4)(1)-(1)(2) = 2, \quad M_{12} = (0)(1)-(1)(5) = -5, \quad M_{13} = (0)(2)-(4)(5) = -20$
Step 3
Combine with signs and coefficients
Multiply each first-row element by its signed minor and sum.
$\det(A) = 2(2) - 1(-5) + 3(-20) = 4 + 5 - 60 = -51$
Problem Statement
Find the inverse of $A = \begin{bmatrix} 3 & 7 \\ 1 & 4 \end{bmatrix}$.
inverse matrixdeterminant
Strategic Toolkit
$$A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
Step 1
Compute the determinant
Check that the determinant is non-zero so the inverse exists.
$\det(A) = (3)(4) - (7)(1) = 12 - 7 = 5$
Step 2
Form the adjugate matrix
Swap the diagonal elements and negate the off-diagonal elements.
$\text{adj}(A) = \begin{bmatrix} 4 & -7 \\ -1 & 3 \end{bmatrix}$
Step 3
Divide by the determinant
Multiply the adjugate by the scalar $\frac{1}{\det(A)}$.
$A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & -7 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 0.8 & -1.4 \\ -0.2 & 0.6 \end{bmatrix}$
Problem Statement
Solve the system of equations: $2x + 3y = 13$ and $x + 4y = 14$.
matrix systemsinverse matrix
Strategic Toolkit
$$\vec{x} = A^{-1}\vec{b}$$
Step 1
Write in matrix form
Extract the coefficient matrix A, variable vector, and constant vector.
$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 13 \\ 14 \end{bmatrix}$
Step 2
Find the determinant
Compute det(A) to verify the system has a unique solution.
$\det(A) = (2)(4) - (3)(1) = 8 - 3 = 5$
Step 3
Find the inverse of A
Apply the 2×2 inverse formula.
$A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
Step 4
Multiply to get the solution
Multiply $A^{-1}$ by the constant vector to find x and y.
$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 13 \\ 14 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 52-42 \\ -13+28 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 10 \\ 15 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$
Memorize formulas
2×2 Determinant
For any 2×2 matrix, subtract the off-diagonal product from the main diagonal product.
Formula
$$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$$
3×3 Determinant (Cofactor Expansion)
Expand along any row or column using signed 2×2 minors with alternating signs.
Formula
$$\det(A) = a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31})$$
2×2 Matrix Inverse
Swap the diagonal, negate the off-diagonal, divide the entire matrix by the determinant.
Formula
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$
Matrix Multiplication Dimension Rule
The product of an m×p and p×n matrix is m×n. Inner dimensions must match.
Formula
$$(m \times \cancel{p})(\cancel{p} \times n) = m \times n$$
Matrix System Solution
Multiply both sides of $A\vec{x} = \vec{b}$ by $A^{-1}$ to isolate the unknown vector.
Formula
$$A\vec{x} = \vec{b} \implies \vec{x} = A^{-1}\vec{b}$$
Review
1
A matrix has 3 rows and 4 columns. How many elements does it contain?
Ans
12
Review
Tap to reveal
Apply the element count formula
Multiply the number of rows by the number of columns.
$$3 \times 4 = 12 \text{ elements}$$
2
If $A = \begin{pmatrix} 2 & 5 \\ 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & -1 \\ 0 & 7 \end{pmatrix}$, find $A + B$.
Ans
$\begin{pmatrix} 6 & 4 \\ 3 & 8 \end{pmatrix}$
Review
Tap to reveal
Add corresponding elements
Add each element of $A$ to the element in the same position of $B$.
$$A + B = \begin{pmatrix} 2+4 & 5+(-1) \\ 3+0 & 1+7 \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 3 & 8 \end{pmatrix}$$
3
If $A = \begin{pmatrix} 1 & -2 \\ 4 & 3 \end{pmatrix}$, what is $-2A$?
Ans
$\begin{pmatrix} -2 & 4 \\ -8 & -6 \end{pmatrix}$
Review
Tap to reveal
Multiply each element by the scalar
Multiply every element of $A$ by $-2$, keeping track of sign changes.
$$-2\begin{pmatrix} 1 & -2 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ -8 & -6 \end{pmatrix}$$
4
The area of a parallelogram spanned by vectors $(3, 5)$ and $(2, 4)$ equals the absolute value of the determinant of the matrix with these vectors as rows. Find the area.
Ans
2
Review
Tap to reveal
Set up the determinant
Arrange the vectors as rows of a $2 \times 2$ matrix and apply the determinant formula $ad - bc$.
$$\det = \begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix} = (3)(4) - (5)(2)$$
Compute and take absolute value
Evaluate the expression and apply the absolute value since area cannot be negative.
$$= 12 - 10 = 2, \qquad \text{Area} = |2| = 2$$
5
Compute the determinant of $\begin{pmatrix} 2 & 0 & 1 \\ 3 & 1 & -1 \\ 0 & 2 & 1 \end{pmatrix}$ using cofactor expansion along the first row.
Ans
12
Review
Tap to reveal
Expand along the first row
Apply cofactor expansion: $\det = a_{11}C_{11} - a_{12}C_{12} + a_{13}C_{13}$.
$$\det = 2\begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} - 0\begin{vmatrix} 3 & -1 \\ 0 & 1 \end{vmatrix} + 1\begin{vmatrix} 3 & 1 \\ 0 & 2 \end{vmatrix}$$
Evaluate each $2 \times 2$ minor
Compute each $2 \times 2$ determinant and combine with the alternating signs.
$$= 2(1+2) - 0 + 1(6-0) = 6 + 6 = 12$$
6
If $P$ is a $4 \times 3$ matrix and the product $PQ$ is a $4 \times 5$ matrix, what are the dimensions of $Q$?
Ans
$3 \times 5$
Review
Tap to reveal
Apply the dimension rule
For the product to be valid, the column count of $P$ must equal the row count of $Q$.
$$(4 \times \cancel{3})(\cancel{3} \times n) = 4 \times 5 \implies n = 5$$
State the dimensions of $Q$
The inner dimension of $Q$ is $3$ and the outer dimension is $5$.
$$Q \text{ is a } 3 \times 5 \text{ matrix}$$
7
A shop tracks quantities purchased by two customers. The quantity matrix is $Q = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ (rows = customers, columns = items) and the price vector is $P = \begin{pmatrix} 10 \\ 5 \end{pmatrix}$ (item 1 costs Rs. 10, item 2 costs Rs. 5). Find $QP$ to get each customer's total bill.
Ans
$\begin{pmatrix} 25 \\ 50 \end{pmatrix}$
Review
Tap to reveal
Set up the matrix multiplication
A $2 \times 2$ matrix times a $2 \times 1$ column vector yields a $2 \times 1$ result.
$$\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 10 \\ 5 \end{pmatrix} = \begin{pmatrix} 2(10)+1(5) \\ 3(10)+4(5) \end{pmatrix}$$
Compute each customer's bill
Evaluate the dot product of each row of $Q$ with the price vector.
$$= \begin{pmatrix} 20+5 \\ 30+20 \end{pmatrix} = \begin{pmatrix} 25 \\ 50 \end{pmatrix}$$
8
A data encoding scheme uses the matrix $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$. Decoding requires computing $A^{-1}$. Find $A^{-1}$.
Ans
$\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$
Review
Tap to reveal
Compute the determinant
Find $\det(A) = ad - bc$ for the $2 \times 2$ matrix.
$$\det(A) = (3)(2) - (1)(5) = 1$$
Apply the inverse formula
Swap $a$ and $d$ on the diagonal, negate $b$ and $c$ off the diagonal, and divide by the determinant.
$$A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$$
9
Solve the system $2x + y = 7$ and $x - y = 2$ using matrix inversion ($A\vec{x} = \vec{b}$).
Ans
$x = 3,\; y = 1$
Review
Tap to reveal
Write the system in matrix form
Express $A\vec{x} = \vec{b}$ with $A$ as the coefficient matrix.
$$\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$$
Find $A^{-1}$ and solve
Compute the inverse and multiply by $\vec{b}$ to get $\vec{x} = A^{-1}\vec{b}$.
$$\vec{x} = \frac{1}{-3}\begin{pmatrix} -1 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 7 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$$
10
A supply-demand model uses the coefficient matrix $\begin{pmatrix} 2 & k \\ 4 & 8 \end{pmatrix}$. The model has no unique solution when this matrix is singular. For what value of $k$ does this occur?
Ans
$4$
Review
Tap to reveal
Set the determinant to zero
A singular matrix has $\det = 0$. Apply the $2 \times 2$ determinant formula.
$$2(8) - k(4) = 0 \implies 16 - 4k = 0$$
Solve for $k$
Isolate $k$ from the linear equation.
$$4k = 16 \implies k = 4$$
11
A bakery produces two items with per-unit costs $C = \begin{pmatrix} 30 & 20 \end{pmatrix}$ (Rs. 30 per unit of item 1, Rs. 20 per unit of item 2). In January they produced $U = \begin{pmatrix} 50 \\ 80 \end{pmatrix}$ units. Find $CU$ (total production cost).
Ans
Rs. 3100
Review
Tap to reveal
Set up the dot product
Multiply the $1 \times 2$ cost row by the $2 \times 1$ quantity column.
$$CU = (30)(50) + (20)(80)$$
Compute the total cost
Evaluate each product and sum.
$$= 1500 + 1600 = \text{Rs. } 3100$$
12
A gift shop sells small boxes for Rs. $x$ and large boxes for Rs. $y$. A customer buys 2 small and 1 large box for Rs. 100. Another buys 1 small and 3 large boxes for Rs. 120. What is the price of a large box?
Ans
Rs. 28
Review
Tap to reveal
Write the system in matrix form
Set up $A\vec{x} = \vec{b}$ with $\vec{x} = (x, y)^T$.
$$\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 100 \\ 120 \end{pmatrix}$$
Compute the inverse and solve
Find $A^{-1}$ and compute $\vec{x} = A^{-1}\vec{b}$.
$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 100 \\ 120 \end{pmatrix} = \begin{pmatrix} 36 \\ 28 \end{pmatrix}$$
13
If $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $I$ is the $2 \times 2$ identity matrix, what is $AI$?
Ans
$\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$
Review
Tap to reveal
Apply the identity matrix property
For any matrix $A$, $AI = IA = A$. The identity matrix has 1s on the diagonal and 0s elsewhere.
$$\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$$
14
The coefficient matrix of a three-variable system is $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & -1 \\ 3 & -1 & 2 \end{pmatrix}$. Find its determinant.
Ans
-28
Review
Tap to reveal
Expand along the first row
Apply cofactor expansion: $\det = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}$.
$$= 1\begin{vmatrix} 1 & -1 \\ -1 & 2 \end{vmatrix} - 2\begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} + 3\begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix}$$
Evaluate and combine
Compute each $2 \times 2$ determinant and apply the alternating signs.
$$= 1(1) - 2(7) + 3(-5) = 1 - 14 - 15 = -28$$
15
If $A = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix}$, what is $AB - BA$?
Ans
$\begin{pmatrix} -2 & -2 \\ 4 & 2 \end{pmatrix}$
Review
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Compute $AB$
Multiply $A$ by $B$ row-by-column.
$$AB = \begin{pmatrix} 1(4)+0(0) & 1(1)+0(2) \\ 2(4)+3(0) & 2(1)+3(2) \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 8 & 8 \end{pmatrix}$$
Compute $BA$
Multiply $B$ by $A$ row-by-column (order matters).
$$BA = \begin{pmatrix} 4(1)+1(2) & 4(0)+1(3) \\ 0(1)+2(2) & 0(0)+2(3) \end{pmatrix} = \begin{pmatrix} 6 & 3 \\ 4 & 6 \end{pmatrix}$$
Find $AB - BA$
Subtract $BA$ from $AB$ element-by-element.
$$AB - BA = \begin{pmatrix} 4-6 & 1-3 \\ 8-4 & 8-6 \end{pmatrix} = \begin{pmatrix} -2 & -2 \\ 4 & 2 \end{pmatrix}$$
16
If $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$, what are the dimensions of $A^T$ (the transpose of $A$)?
Ans
$3 \times 2$
Review
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Identify the original dimensions
Matrix $A$ has 2 rows and 3 columns.
$$A \text{ is } 2 \times 3$$
Apply the transpose
The transpose flips rows and columns: a $2 \times 3$ matrix becomes $3 \times 2$.
$$A^T \text{ is } 3 \times 2$$
17
If the determinant of $\begin{pmatrix} x & 4 \\ 3 & x+2 \end{pmatrix}$ equals $-12$, find the possible values of $x$.
Ans
$x = 0 \text{ or } x = -2$
Review
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Set up the determinant equation
Apply the $2 \times 2$ determinant formula and equate to $-12$.
$$x(x+2) - (4)(3) = -12 \implies x^2 + 2x - 12 = -12$$
Solve the quadratic
Simplify and factor to find the values of $x$.
$$x^2 + 2x = 0 \implies x(x+2) = 0 \implies x = 0 \text{ or } x = -2$$
18
Which of the following statements about matrices is always true?
Ans
For a square matrix $A$, $\det(A) = \det(A^T)$
Review
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Evaluate each statement
Check each claim against known matrix properties.
$$\det(A) = \det(A^T) \text{ is always true for square matrices}$$
19
The sum of two monthly sales matrices is $\begin{pmatrix} 5 & 3 \\ 7 & 1 \end{pmatrix}$ (in thousands of rupees) and their difference (Month 1 minus Month 2) is $\begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix}$. Find the sales matrix for Month 1.
Ans
$\begin{pmatrix} 3 & 1 \\ 5 & -1 \end{pmatrix}$
Review
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Set up the system of matrix equations
Let $A$ be Month 1 and $B$ be Month 2. We have $A + B$ and $A - B$.
$$A + B = \begin{pmatrix} 5 & 3 \\ 7 & 1 \end{pmatrix}, \qquad A - B = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix}$$
Solve for $A$ by adding the equations
Adding $A + B$ and $A - B$ eliminates $B$, giving $2A$.
$$2A = \begin{pmatrix} 6 & 2 \\ 10 & -2 \end{pmatrix} \implies A = \begin{pmatrix} 3 & 1 \\ 5 & -1 \end{pmatrix}$$
20
A company has two factories producing three products. The daily output matrix is $P = \begin{pmatrix} 20 & 15 & 10 \\ 25 & 10 & 20 \end{pmatrix}$ (rows: factories, columns: products). Selling prices per unit are $S = \begin{pmatrix} 100 \\ 200 \\ 150 \end{pmatrix}$. Find the total daily revenue from both factories combined.
Ans
Rs. 14000
Review
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Compute the revenue per factory
Multiply the output matrix $P$ by the price vector $S$.
$$PS = \begin{pmatrix} 20(100)+15(200)+10(150) \\ 25(100)+10(200)+20(150) \end{pmatrix} = \begin{pmatrix} 6500 \\ 7500 \end{pmatrix}$$
Sum the factory revenues
Add the revenues from both factories to get the total daily revenue.
$$6500 + 7500 = \text{Rs. } 14000$$
Question 1 of 20
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1
A matrix has 3 rows and 4 columns. How many elements does it contain?