Matrices and Linear Equations: Exercises with Worked Solutions

These ten exercises follow the companion guide on matrices and linear equations. They run from vector equality and the dot product through matrix multiplication, determinants, cofactors, and the inverse, finishing with all three methods for solving a linear system. Work each one fully before turning to the solutions, which explain the reasoning step by step rather than just stating an answer.

Exercises

Exercise 1. Decide whether each pair of vectors is equal, and justify your answer.

Exercise 2. Given the two row vectors below, compute the transpose of the first, the scalar multiple two times the first, their sum, and their difference.

Exercise 3. A market stall sells three items at prices of 60, 45, and 80 pence. In one hour it sells 4, 6, and 2 of them respectively. Using a dot product, find the total revenue and state it in pounds.

Exercise 4. Consider the matrix below. State its dimension, say whether it is square, and write down the entries in row 2 column 4 and in row 3 column 1.

Exercise 5. For the two matrices below, find the transpose of the first, then compute the linear combination three times the first minus two times the second.

Exercise 6. For the matrices below, compute the product of the first times the second, and explain whether the product in the opposite order exists.

Exercise 7. Compute the determinant of the matrix below using cofactor expansion along the first row.

Exercise 8. For the same matrix as Exercise 7, find the minor and the cofactor of the entry in row 2 column 3.

Exercise 9. Find the inverse of the matrix below, then use it to solve the system where the right-hand side is the vector with entries 4 and 9.

Exercise 10. Solve the following system using Cramer’s rule.

Solutions

Solution 1. Two vectors are equal only when they share a dimension and agree in every component. In part (a) the dimensions match and every corresponding entry is identical, so the vectors are equal. In part (b) the dimensions match but the first entries already disagree, three is not one, so the vectors are not equal. Order matters.

Solution 2. Transposing turns the row into a column.

Scalar multiplication scales every entry.

Addition and subtraction work entry by entry.

Solution 3. The dot product of the price vector with the quantity vector multiplies corresponding entries and sums them.

The total revenue is 670 pence, which is six pounds seventy.

Solution 4. The matrix has three rows and four columns, so its dimension is three by four. Since the number of rows differs from the number of columns it is not square. The entry in row 2 column 4 is 6, and the entry in row 3 column 1 is 8.

Solution 5. The transpose turns each row into a column.

Scaling each matrix first gives the two pieces of the combination.

Subtracting entry by entry gives the result.

Solution 6. The first matrix is two by two and the second is two by three. The inner dimensions agree, both two, so the product exists and is two by three. Each entry is the dot product of a row of the first with a column of the second.

The product in the opposite order does not exist, because the second matrix has three columns while the first has only two rows, and multiplication requires those two counts to match.

Solution 7. Expanding along the first row, the third term vanishes because its coefficient is zero, leaving two two-by-two determinants to evaluate.

Evaluating the two minors gives six and minus two.

Solution 8. Deleting row 2 and column 3 leaves the two-by-two block formed by rows 1 and 3 and columns 1 and 2.

Its determinant is the minor.

Applying the sign factor, where two plus three is odd, flips the sign.

Solution 9. For a two by two matrix, the determinant is the product of the diagonal minus the product of the off-diagonal.

Because the determinant is nonzero the inverse exists. Swap the diagonal entries, negate the off-diagonal ones, and divide by the determinant.

The solution is the inverse times the right-hand side.

So the solution is x equals 3 and y equals minus 2, which you can confirm by substituting back into the original two equations.

Solution 10. Write the coefficient matrix and the right-hand side.

Cofactor expansion along the first row gives the determinant of the coefficient matrix.

Replacing the first column with the right-hand side and taking the determinant gives the numerator for the first unknown.

Replacing the second column with the right-hand side gives the numerator for the second unknown.

Replacing the third column gives the numerator for the third unknown.

Each unknown is its numerator divided by the determinant of the coefficient matrix.

The solution is x equals 1, y equals 1, and z equals 1, which checks against all three original equations.

See you soon.