Matrices: Application Problems (5)

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Question

Solve the system using matrix inversion:

2x+3y−z=5,x−2y+3z=−2,3x+y+2z=72x + 3y - z = 5,\quad x - 2y + 3z = -2,\quad 3x + y + 2z = 7

Solution — Step by Step

A=(23−11−23312),X=(xyz),B=(5−27)A = \begin{pmatrix} 2 & 3 & -1 \\ 1 & -2 & 3 \\ 3 & 1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ -2 \\ 7 \end{pmatrix}

Expanding along row 1:

det⁡A=2[(−2)(2)−(3)(1)]−3[(1)(2)−(3)(3)]+(−1)[(1)(1)−(−2)(3)]\det A = 2[(-2)(2) - (3)(1)] - 3[(1)(2) - (3)(3)] + (-1)[(1)(1) - (-2)(3)] =2(−4−3)−3(2−9)−1(1+6)=−14+21−7=0= 2(-4 - 3) - 3(2 - 9) - 1(1 + 6) = -14 + 21 - 7 = 0

Determinant is zero — so AA is singular and A−1A^{-1} does not exist!

Since det⁡A=0\det A = 0, the system either has no solution or infinitely many. Use elementary row reduction on the augmented matrix:

(23−1∣51−23∣−2312∣7)→R3→R1+R2(23−1∣51−23∣−2312∣3)\begin{pmatrix} 2 & 3 & -1 & | & 5 \\ 1 & -2 & 3 & | & -2 \\ 3 & 1 & 2 & | & 7 \end{pmatrix} \xrightarrow{R_3 \to R_1 + R_2} \begin{pmatrix} 2 & 3 & -1 & | & 5 \\ 1 & -2 & 3 & | & -2 \\ 3 & 1 & 2 & | & 3 \end{pmatrix}

Wait — R1+R2=(3,1,2∣3)R_1 + R_2 = (3, 1, 2 \mid 3), but the actual R3R_3 has constant 77, not 33. So R1+R2≠R3R_1 + R_2 \ne R_3 on the constant side, while R3=R1+R2R_3 = R_1 + R_2 on the coefficient side.

The coefficient matrix has rank 2, but the augmented matrix has rank 3 (the constants column is inconsistent with the linear dependence among rows). The system is inconsistent — no solution exists.

Why This Works

Matrix inversion only solves a system when det⁡A≠0\det A \ne 0 (the system has a unique solution). When det⁡A=0\det A = 0, we must check rank to distinguish “no solution” from “infinitely many solutions.”

The diagnostic is: compare rank of AA with rank of [A∣B][A | B]. Equal ranks → solutions exist (infinitely many, with the parameter count = nn − rank). Unequal ranks → no solution.

Alternative Method

Cramer’s rule with det⁡A=0\det A = 0 also signals trouble — but cannot distinguish “no solution” from “infinite solutions.” Row reduction is the most reliable approach when det⁡A=0\det A = 0.

Common Mistake

Students who notice det⁡A=0\det A = 0 sometimes panic and write ”AA is singular, problem has no solution.” Wrong — it could have infinitely many. Always check the augmented-matrix rank to decide. Conversely, students who never check the determinant blindly try X=A−1BX = A^{-1}B, get nonsense, and lose all marks.

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