Step 1

Let A, B, and C be square matrices such that AB=AC.

If \(A \neq 0\), then B=C.

Determine true or false.

Step 2

Let A, B, and C be square matrices such that AB=AC.

If \(A \neq 0\) , then B=C.

This is true only if matrix A is invertible.

If A is invertible then \(A^{-1}\) exist.

Multiply given equationAB=AC by \(A^{-1}\) on both sides,

\(A^{-1}AB=A^{-1}AC\)

\((A^{-1}A)B=(A^{-1}A)C\)

\(I \times B = I \times C\)

B=C

If A is not invertible then this statement is false.

Counter example:

\(A=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix} , B=\begin{bmatrix}8 & 4 \\5 & 5 \end{bmatrix} , C=\begin{bmatrix}5 & -2 \\3 & 1 \end{bmatrix}\)

Here, \(det|A|=2 \times 6 -(-4) \times (-3)\)

=12-12

=0

That is A is not invertible.

Now, find AB and AC,

\(AB=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix}\begin{bmatrix}8 & 4 \\5 & 5 \end{bmatrix}\)

\(=\begin{bmatrix}16-15 & 8-15 \\-32+30 & -16+30 \end{bmatrix}\)

\(=\begin{bmatrix}1 & -7 \\-2 & 14 \end{bmatrix}\)

\(AC=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix}\begin{bmatrix}5 & -2 \\3 & 1 \end{bmatrix}\)

\(=\begin{bmatrix}10-9 & -4-3 \\-20+18 & 8+6 \end{bmatrix}\)

\(=\begin{bmatrix}1 & -7 \\-2 & 14 \end{bmatrix}\)

That is, AB=AC but B is not the same as C.

Let A, B, and C be square matrices such that AB=AC.

If \(A \neq 0\), then B=C.

Determine true or false.

Step 2

Let A, B, and C be square matrices such that AB=AC.

If \(A \neq 0\) , then B=C.

This is true only if matrix A is invertible.

If A is invertible then \(A^{-1}\) exist.

Multiply given equationAB=AC by \(A^{-1}\) on both sides,

\(A^{-1}AB=A^{-1}AC\)

\((A^{-1}A)B=(A^{-1}A)C\)

\(I \times B = I \times C\)

B=C

If A is not invertible then this statement is false.

Counter example:

\(A=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix} , B=\begin{bmatrix}8 & 4 \\5 & 5 \end{bmatrix} , C=\begin{bmatrix}5 & -2 \\3 & 1 \end{bmatrix}\)

Here, \(det|A|=2 \times 6 -(-4) \times (-3)\)

=12-12

=0

That is A is not invertible.

Now, find AB and AC,

\(AB=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix}\begin{bmatrix}8 & 4 \\5 & 5 \end{bmatrix}\)

\(=\begin{bmatrix}16-15 & 8-15 \\-32+30 & -16+30 \end{bmatrix}\)

\(=\begin{bmatrix}1 & -7 \\-2 & 14 \end{bmatrix}\)

\(AC=\begin{bmatrix}2 & -3 \\-4 & 6 \end{bmatrix}\begin{bmatrix}5 & -2 \\3 & 1 \end{bmatrix}\)

\(=\begin{bmatrix}10-9 & -4-3 \\-20+18 & 8+6 \end{bmatrix}\)

\(=\begin{bmatrix}1 & -7 \\-2 & 14 \end{bmatrix}\)

That is, AB=AC but B is not the same as C.