Question:
If
\[
A=
\begin{bmatrix}
c & -a & b\\
a & b & -c\\
-b & c & a
\end{bmatrix},
\qquad
B=
\begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 2\\
1 & 2 & 0
\end{bmatrix}
\]
and
\[
AB=
\begin{bmatrix}
4 & 4 & -2\\
1 & 1 & 10\\
-1 & 5 & -4
\end{bmatrix},
\]
then \(a^2+b^2+c^2=\)
If
\[
A=
\begin{bmatrix}
c & -a & b\\
a & b & -c\\
-b & c & a
\end{bmatrix},
\qquad
B=
\begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 2\\
1 & 2 & 0
\end{bmatrix}
\]
and
\[
AB=
\begin{bmatrix}
4 & 4 & -2\\
1 & 1 & 10\\
-1 & 5 & -4
\end{bmatrix},
\]
then \(a^2+b^2+c^2=\)
Show Hint
In matrix multiplication problems with unknown entries, compare corresponding entries of the product matrix. Often only one row or one column is sufficient to determine all unknowns.
Updated On: Jul 29, 2026
- \(14\)
- \(17\)
- \(11\)
- \(19\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
When two matrices satisfy \(AB=C\), corresponding entries of the product matrix can be equated to obtain equations in the unknowns.
Step 1: Compute the product \(AB\). The first row of \(AB\) is \[ \bigl[c,\,-a,\,b\bigr] \begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & 2\\ 1 & 2 & 0 \end{bmatrix} = \bigl[c+b,\,-a+2b,\,2c-2a\bigr]. \] Comparing with the first row of the given matrix, \[ [c+b,\,-a+2b,\,2c-2a]=[4,\,4,\,-2]. \] Hence, \[ c+b=4, \] \[ -a+2b=4, \] \[ c-a=-1. \]
Step 2: Solve for \(a,b,c\). From \[ c-a=-1, \] we get \[ c=a-1. \] Using \(c+b=4\), \[ a-1+b=4 \] \[ a+b=5. \] Also, \[ -a+2b=4. \] Solving \[ a+b=5, \] \[ -a+2b=4, \] we obtain \[ 3b=9 \] \[ b=3. \] Therefore, \[ a=2 \] and \[ c=1. \]
Step 3: Find \(a^2+b^2+c^2\). Substituting the values, \[ a^2+b^2+c^2=2^2+3^2+1^2. \] \[ =4+9+1. \] \[ =14. \] Therefore, \[ \boxed{a^2+b^2+c^2=14} \]
Step 1: Compute the product \(AB\). The first row of \(AB\) is \[ \bigl[c,\,-a,\,b\bigr] \begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & 2\\ 1 & 2 & 0 \end{bmatrix} = \bigl[c+b,\,-a+2b,\,2c-2a\bigr]. \] Comparing with the first row of the given matrix, \[ [c+b,\,-a+2b,\,2c-2a]=[4,\,4,\,-2]. \] Hence, \[ c+b=4, \] \[ -a+2b=4, \] \[ c-a=-1. \]
Step 2: Solve for \(a,b,c\). From \[ c-a=-1, \] we get \[ c=a-1. \] Using \(c+b=4\), \[ a-1+b=4 \] \[ a+b=5. \] Also, \[ -a+2b=4. \] Solving \[ a+b=5, \] \[ -a+2b=4, \] we obtain \[ 3b=9 \] \[ b=3. \] Therefore, \[ a=2 \] and \[ c=1. \]
Step 3: Find \(a^2+b^2+c^2\). Substituting the values, \[ a^2+b^2+c^2=2^2+3^2+1^2. \] \[ =4+9+1. \] \[ =14. \] Therefore, \[ \boxed{a^2+b^2+c^2=14} \]
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