Matrix Identities: Finding a Value and Proving a Cubic Relation

This lesson solves two exercise questions on matrix powers. The first finds an unknown constant by comparing two matrices entry by entry, and the second proves a 3 by 3 matrix satisfies a cubic identity.

Question 17: solving for $k$

We are given

$A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \qquad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

and told that $A^2 = kA - 2I$. We need the value of $k$.

Step 1: compute $A^2$

Multiply $A$ by itself using the row into column rule:

$A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 9 - 8 & -6 + 4 \\ 12 - 8 & -8 + 4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$

Step 2: build $kA - 2I$

$kA - 2I = k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$

Step 3: compare entries

Setting $A^2 = kA - 2I$ and matching corresponding entries:

$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$

From the top left, $3k - 2 = 1$, so $3k = 3$ and $k = 1$. The other entries confirm it: $-2k = -2$, $4k = 4$, and $-2k - 2 = -4$ all give $k = 1$.

$k = 1$

Second example: a 3 by 3 identity

Given

$A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix},$

show that $A^3 - 23A - 40I = 0$.

Step 1: find $A^2$

$A^2 = A \cdot A = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}$

Be careful to copy each entry exactly when you carry it forward.

Step 2: find $A^3$

Multiply $A$ by $A^2$:

$A^3 = A \cdot A^2 = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}\begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix} = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}$

Step 3: substitute and subtract

Now form $A^3 - 23A - 40I$:

$\begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix} - 23\begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix} - 40\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

The scaled matrices are

$23A = \begin{bmatrix} 23 & 46 & 69 \\ 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix}, \qquad 40I = \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 40 \end{bmatrix}$

Subtracting entry by entry, for example the top left is $63 - 23 - 40 = 0$, and every other entry also cancels:

$A^3 - 23A - 40I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Hence proved.

Key takeaways

• To find an unknown in a matrix equation, compute each side and match corresponding entries; any one matching entry gives the value, and the rest confirm it.
• Multiply matrices with the row into column rule, and copy entries carefully when reusing a result.
• Build a cube as $A \cdot A^2$, then substitute into the identity; if the matrix satisfies the relation, all entries subtract to the zero matrix.