Practice: Determinants

Recognition · Interpretation

A $3 \times 3$ matrix $A$ has $det A = - 21$ . What is $det (3 A)$ ?

2 hints available, least help first.

Hint 1: Retrieval cue

How many rows does scaling the whole matrix by 3 affect?

Hint 2: Concept cue

Each scaled row contributes one factor of 3 to the determinant.

Direct application

For

A = (\begin{matrix} 2 & - 1 & 3 \\ 1 & 4 & - 2 \\ 3 & 0 & 1 \end{matrix}),

expanding along the first row, what is the term contributed by the entry $a_{12} = - 1$ ?

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

What is $(- 1)^{1 + 2}$ , and what minor is left after deleting row 1 and column 2?

Hint 2: Concept cue

The minor is $(\begin{matrix} 1 & - 2 \\ 3 & 1 \end{matrix})$ . Compute its determinant, then multiply by the entry and by the position sign.

Direct application · Prediction

A $3 \times 3$ matrix $B$ has $det B = 6$ . Three operations are applied in turn:

rows 1 and 3 are swapped;
row 2 is multiplied by 4;
twice row 1 is added to row 3.

What is the determinant of the resulting matrix?

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Which of the three operations leaves a determinant unchanged?

Hint 2: Concept cue

Work through them one at a time, updating the value after each.

Recognition · Interpretation

A $3 \times 3$ matrix $M$ is found to have $det M = 0$ . Which statement follows?

2 hints available, least help first.

Hint 1: Retrieval cue

What does a nonzero determinant certify, and what therefore fails when it is zero?

Hint 2: Concept cue

Relate the verdict to the columns, to the rank, and to what the map sends to zero.

Recognition · Error diagnosis

A learner asserts that $det (A + B) = det A + det B$ and offers $A = B = I_{2}$ as a check, reporting $det A + det B = 1 + 1 = 2$ .

Which response identifies the error?

2 hints available, least help first.

Hint 1: Retrieval cue

Compute $A + B$ first, then take its determinant, rather than taking the determinants first.

Hint 2: Concept cue

$A + B = 2 I_{2}$ . Evaluate $det (\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix})$ directly and compare with $2$ .

Construction · Direct application · Explanation

(a) Compute $det A$ for

A = (\begin{matrix} 4 & 0 & 0 & 0 \\ 7 & - 2 & 0 & 0 \\ 1 & 5 & 3 & 0 \\ 9 & 6 & 8 & - 1 \end{matrix}),

saying which method you chose and why the alternative would have been more work.

(b) Compute $det B$ for

B = (\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 5 & 0 & 1 \end{matrix}),

and state everything the value tells you about $B$ and about the map $x \mapsto B x$ .

(c) Compute $det C$ for

C = (\begin{matrix} 2 & - 1 & 0 \\ 3 & 1 & 4 \\ 0 & 5 & - 2 \end{matrix})

by expanding along the row or column that minimises work, stating which you chose and why.

(d) Using your answers, give $det (B C)$ and $det (2 C)$ without computing either product or scaling.

(e) Interpret $det C$ geometrically: what happens to volume, and what happens to orientation?

Write your answer

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

Before computing anything, look for structure: is the matrix triangular, or is one row a multiple of another?

Hint 2: Concept cue

Expand along the row or column containing a zero, that term vanishes and the minor need not be formed.

Hint 3: Strategy cue

For part (d), use $det (X Y) = det X det Y$ and $det (k X) = k^{n} det X$ rather than forming either matrix.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) $det A = 24$ . $A$ is lower triangular, every entry above the diagonal is zero, so the determinant is the product of the diagonal entries:

det A = 4 \times (- 2) \times 3 \times (- 1) = 24 .

Method: recognition, requiring three multiplications. Cofactor expansion on a $4 \times 4$ would generate up to 24 terms and four $3 \times 3$ minors. Expanding along the first row would in fact collapse quickly here, since only $a_{11}$ is nonzero, and recursing that way reproduces the diagonal product, which is one way to see why the triangular rule holds. (b) $det B = 0$ . Row 2 is exactly twice row 1: $(2, 4, 6) = 2 (1, 2, 3)$ . A matrix with one row a multiple of another is singular, so the determinant is zero with no arithmetic needed. Confirming by expansion along column 2, whose entries are $2, 4, 0$ :

- 2 det (\begin{matrix} 2 & 6 \\ 5 & 1 \end{matrix}) + 4 det (\begin{matrix} 1 & 3 \\ 5 & 1 \end{matrix}) - 0 = - 2 (2 - 30) + 4 (1 - 15) = 56 - 56 = 0 ✓

What it tells us: $B$ is not invertible; its columns are linearly dependent; its rank is less than 3 (in fact 2, since rows 1 and 3 are independent); the map $x \mapsto B x$ has a nontrivial kernel, so by rank–nullity its nullity is 1 and its image is a plane rather than all of $R^{3}$ ; and $B x = b$ has either no solution or infinitely many, never exactly one. (c) $det C = - 50$ . Row 1 and column 1 and column 3 each contain one zero, so any is a reasonable choice. Expanding along row 1, whose entries are $2, - 1, 0$ , kills the third term:

det C = + 2 det (\begin{matrix} 1 & 4 \\ 5 & - 2 \end{matrix}) - (- 1) det (\begin{matrix} 3 & 4 \\ 0 & - 2 \end{matrix}) + 0 .

The first minor: $1 (- 2) - 4 (5) = - 22$ . The second: $3 (- 2) - 4 (0) = - 6$ .

det C = 2 (- 22) + 1 (- 6) = - 44 - 6 = - 50 .

Checking by expansion along column 1, entries $2, 3, 0$ :

+ 2 det (\begin{matrix} 1 & 4 \\ 5 & - 2 \end{matrix}) - 3 det (\begin{matrix} - 1 & 0 \\ 5 & - 2 \end{matrix}) + 0 = 2 (- 22) - 3 (2) = - 44 - 6 = - 50 ✓

So $det C = - 50$ . Both routes cost two $2 \times 2$ determinants because each chosen line had a zero; expanding along row 2 or column 2, which have no zeros, would have cost three. (d) Using multiplicativity and the scaling rule.

det (B C) = det B \cdot det C = 0 \times (- 50) = 0 .

The product is singular, as it must be: composing with a map that collapses a direction cannot recover it.

det (2 C) = 2^{3} det C = 8 (- 50) = - 400,

since $C$ is $3 \times 3$ and scaling the matrix scales all three rows. (e) Geometric reading of $det C = - 50$ . The magnitude says the map multiplies every volume by 50: the unit cube becomes a parallelepiped of volume 50, and any region's volume is scaled by the same factor. The negative sign says orientation is reversed. A right-handed frame is carried to a left-handed one, so the map is not a rotation or a stretch but includes a reflection. Since $det C \neq 0$ , nothing is collapsed: the map is invertible, its columns are independent, and its inverse has determinant $- 1 / 50$ .

A complete answer does each of these:

applies cofactor signs
tracks row operations
chooses an efficient route
reads the verdict
interprets geometrically

Practice: Determinants

Recognition · Interpretation

Direct application

Direct application · Prediction

Recognition · Interpretation

Recognition · Error diagnosis

Construction · Direct application · Explanation

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